Additional Question (Limit)[Page 392]

1. Prove that \( \displaystyle \lim_{x \to \frac{2}{3}} \frac{2}{2-3x}\) does NOT exist.

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to \frac{2}{3}} \frac{2}{2-3x}\)
   
   At \(x= \frac{2}{3}\), we compute the following
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to \frac{2}{3}^-} f(x)= \lim_{h \to 0^-} \frac{2}{2-3(\frac{2}{3}-h)}=\lim_{h \to 0^-}\frac{2}{2-2+h}=\lim_{h \to 0^-}\frac{2}{h}=\) ∞
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to \frac{2}{3}^+} f(x)= \lim_{h \to 0^+} \frac{2}{2-3(\frac{2}{3}+h)}=\lim_{h \to 0^+}\frac{2}{2-2-h}=\lim_{h \to 0^+}\frac{2}{-h}=\) -∞
   
   Since,
   LHL≠RHL
   
   The limit does NOT exist at \(x= \frac{2}{3}\)

   
   This completes the solution
2. Do the following function define for the value x=1?
   1. \( f(x)=\frac{x-1}{x+2}\)
      
      

      Solution 👉 Click Here

      Yes, the function \( f(x)=\frac{x-1}{x+2}\) is defined for the value x=1, because when we put x=1 to the function f(x), then
      
      \( f(x)=\frac{x-1}{x+2}\)
      
      = \( f(x)=\frac{1-1}{1+2}\)
      
      = \( f(x)=\frac{0}{3}\)
      
      = \( f(x)=0\)
      
      
   2. \( f(x)=\frac{x^3+1}{x-1}\)

      Solution 👉 Click Here

      No, the function \( f(x)=\frac{x^3+1}{x-1}\) is NOT defined for the value x=1, because when we put x=1 to the function f(x), then
      
      \( f(x)=\frac{x^3+1}{x-1}\)
      
      = \( f(x)=\frac{1^3+1}{1-1}\)
      
      = \( f(x)=\frac{2}{0}\)
      
      = \( f(x)=\infty\)
      
      
3. What do you mean by the left hand limit and right hand limit of a function? What is the condition for the limit of a function to exist at a point?
   Prove that \(\displaystyle \lim_{x \to 0}|x|=0\) but \(\displaystyle \lim_{x \to 0} \frac{|x|}{x} \) does not exist.

   Solution 👉 Click Here

   Left-hand limit

   The left-hand limit of a function f(x) at a particular point x=a is used to determine the behavior of the function f(x) as it approaches a from the left. Left-Hand Limit (LHL): is defined as
   \( LHL=\displaystyle \lim_{x \to a^-} f(x)\)
   
   This limit represents the behavior of the function f(x) as x gets closer to a while staying to the left of a.

   Right-hand limit

   The right-hand limit of a function f(x) at a particular point x=a is used to determine the behavior of the function f(x) as it approaches a from the right. Right-Hand Limit (RHL): is defined as
   \( RHL=\displaystyle \lim_{x \to a^+} f(x)\)
   
   This limit represents the behavior of the function f(x) as x gets closer to a while staying to the right of a.

   Condition for the limit of a function to exist at a point

   let f(x) be a function, then condition for the limit of a function f(x) to exist at a point a is that
   \( LHL=RHL\)
   
   or \( \displaystyle \lim_{x \to a^-} f(x)=\lim_{x \to a^+} f(x)\)
   
   This limit represents the behavior of the function f(x) as x gets closer to the point a, from both sides:left and right.

   \(\displaystyle \lim_{x \to 0}|x|=0\)

   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 0^-} |x|= \lim_{x \to 0^-} -x = \) 0
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 0^+} |x|= \lim_{x \to 0^+} x = \) 0
   
   Since,
   LHL=RHL
   
   The limit exist at x=0, and the limit value is 0

   \(\displaystyle \lim_{x \to 0} \frac{|x|}{x} \) does NOT exist

   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 0^-} \frac{|x|}{x}= \lim_{x \to 0^-} \frac{-x}{x} = \) -1
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 0^+} \frac{|x|}{x}= \lim_{x \to 0^+} \frac{x}{x} = \) 1
   
   Since,
   LHL≠RHL
   
   The limit does NOT exist at x=0

4. Distinguish between limit and value of a function.
   It is given that \(f(x)=\frac{ax+b}{x+1},\displaystyle \lim_{x \to 0} f(x)=2\) and \( \displaystyle \lim_{x \to \infty} f(x)=1\). Prove that \(f(-2)=0\)

   Solution 👉 Click Here

   Distinguish between limit and value

   	Limit of a Function	Value of a Function
   Behaviour at a	represents the behavior of the function as it gets closer and closer to a particular point (typically denoted as "a")	The value of a function represents the actual output or result of the function when a specific input is provided.
   Symbolically	the limit of a function f(x) as x approaches a is denoted as \( \displaystyle \lim_{x \to a} f(x) \)	the value of a function f(x) at a particular point a is denoted as f(a)
   Definition at a	The limit may not necessarily be equal to the function's actual value at that point. It describes what happens to the function's values as x gets arbitrarily close to a.	It provides information about the function's behavior at a specific point.
   	Limits are used to analyze the behavior of functions near singular/critical points, to determine continuity, derivatives and integrals.	Functional value are used to compute the value at a

   Problem Solving

   It is given that \(f(x)=\frac{ax+b}{x+1}\)
   
   Also given that
   \(\displaystyle \lim_{x \to 0} f(x)=2\)
   
   or\(\displaystyle \lim_{x \to 0} \frac{ax+b}{x+1}=2\)
   
   or\( \frac{a.0+b}{0+1}=2\)
   
   or\( b=2\)
   
   Also given that
   \( \displaystyle \lim_{x \to \infty} f(x)=1\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{ax+b}{x+1}=1\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{a+\frac{b}{x}}{1+\frac{1}{x}}=1\)
   
   or\( \frac{a+\frac{b}{\infty}}{1+\frac{1}{\infty}}=1\)
   
   or\( \frac{a+0}{1+0}=1\)
   
   or\( a=1\)
   
   Now
   \(f(x)=\frac{ax+b}{x+1}\)
   
   or\(f(-2)=\frac{a.(-2)+b}{(-2)+1}\)
   
   or\(f(-2)=\frac{1.(-2)+2}{(-2)+1}\)
   
   or\(f(-2)=0\)
   
   This completes the proof.

5. Define limit of a function at a point. It is given that \(f(x)=\frac{x+6}{cx-d},\displaystyle \lim_{x \to 0} f(x)=-6 \) and \( \displaystyle \lim_{x \to \infty} f(x)=\frac{1}{3}\).
   Prove that \(f(13)=\frac{1}{2}\)

   Solution 👉 Click Here

   Limit of a function

   Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
   Then we say that \( \displaystyle \lim_{x \to a} f(x) =L\) if for a given number \(\epsilon > 0\), there exists a number \(\delta > 0\)such that
   |f(x)-L|<\(\epsilon\) whenever |x-a|<\(\delta\)

   Empirical Definition of Limit

   Let 𝑓(𝑥) be a function defined at all values in an open interval containing a, with the possible exception of "a" itself, and let L be a real number.
   Then we say that \( \displaystyle \lim_{x \to a} f(x) =L\) if all of the following three conditions hold

   1. \( \displaystyle \lim_{x \to a^{-}} f(x) \) exists=LHS
   2. \( \displaystyle \lim_{x \to a^{+}} f(x) \) exists=RHS
   3. LHS=RHS

   Problem Solving

   It is given that \(f(x)=\frac{x+6}{cx-d}\)
   
   Also given that
   \(\displaystyle \lim_{x \to 0} f(x)=-6\)
   
   or\(\displaystyle \lim_{x \to 0} \frac{x+6}{cx-d}=-6\)
   
   or\( \frac{0+6}{c.0-d}=-6\)
   
   or\( d=1\)
   
   Also given that
   \( \displaystyle \lim_{x \to \infty} f(x)=\frac{1}{3}\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{x+6}{cx-d}=\frac{1}{3}\)
   
   or\( \displaystyle \lim_{x \to \infty} \frac{1+\frac{6}{x}}{c-\frac{d}{x}}=\frac{1}{3}\)
   
   or\( \frac{1+0}{c-0}=\frac{1}{3}\)
   
   or\( \frac{1}{c}=\frac{1}{3}\)
   
   or\( c=3\)
   
   Now
   \(f(x)=\frac{x+6}{cx-d}\)
   
   or\(f(13)=\frac{13+6}{c.13-d}\)
   
   or\(f(13)=\frac{13+6}{3.13-1}\)
   
   or\(f(13)=\frac{1}{2}\)
   
   This completes the proof.

6. What do you mean by an indeterminate form? State their different forms. Evaluate the following limit \(\displaystyle \lim_{x \to \infty} \sqrt{x} (\sqrt{x}-\sqrt{x-a})\)

   Solution 👉 Click Here

   Indeterminate form

   The term "indeterminate" refers to a situation where the value of an expression cannot be uniquely determined based solely on its appearance.

   The different form of such indeterminates are as follows.

   1. \(\frac{0}{0}\)
      

      The expression "\(\frac{0}{0}\)" is called indeterminate because it doesn't provide enough information to uniquely determine its value

      For example

      \(\frac{1}{1}=1\)	\(\frac{1}{0}=\infty\)	\(\frac{0}{1}=0\)
      \(\frac{2}{2}=1\)	\(\frac{2}{0}=\infty\)	\(\frac{0}{2}=0\)
      \(\frac{3}{3}=1\)	\(\frac{3}{0}=\infty\)	\(\frac{0}{3}=0\)
      …	…	…
      \(\frac{a}{a}=1\)	\(\frac{a}{0}=\infty\)	\(\frac{0}{a}=0\)
      \(\frac{0}{0}=1\)	\(\frac{0}{0}=\infty\)	\(\frac{0}{0}=0\)

      Here, \(\frac{0}{0}\) creates a situation where there is uncertainty about how the fraction \(\frac{0}{0}\) as a whole behaves. In other words, knowing that both the numerator and denominator are approaching zero doesn't immediately mean \(\frac{0}{0}\) will approach a specific finite value, approach infinity, or approach zero. So, it is indeterminate

   2. \(\frac{\infty}{\infty}\)
      Usually \(\frac{\infty}{number}=\infty\) and \(\frac{number}{\infty}=0\). So the top pulls the limit up to infinity and the bottom tries to pull it down to 0. So who wins?
   3. \(0.\infty\)
      Usually 0 · (number) = 0 and (number) · ∞ = ∞. So one piece tries to pull the limit down to zero, and the other tries to pull it up to ∞. Does one side win?
   4. \(\infty-\infty\)
      In general ∞ − (number) = ∞, but (number) − ∞ = −∞. So who wins?
   5. \(\infty^{0}\)
      In general ∞ raised to any positive power should be equal to ∞, ∞ raised to a negative power is 0, and anything raised to the zero should be equal to 1. So who wins?
   6. \(1^{\infty}\)
      Usually 1 raised to any power is just equal to 1. But fractions raised to the ∞ goes to zero, and numbers larger than 1 raised to the ∞ should go off to ∞. So where does \(1^{\infty}\) go?
   7. \(0^{0}\)
      In general zero raised to any positive power is just zero, but but anything raised to the zero should be equal to 1. So which is it?

   Problem Solving

   \(\displaystyle \lim_{x \to \infty} \sqrt{x} (\sqrt{x}-\sqrt{x-a})\)	= \(\displaystyle \lim_{x \to \infty} \sqrt{x} (\sqrt{x}-\sqrt{x-a}) \frac{(\sqrt{x}+\sqrt{x-a})}{(\sqrt{x}+\sqrt{x-a})}\)
   	=\(\displaystyle \lim_{x \to \infty} \sqrt{x} (x-x+a) \frac{1}{(\sqrt{x}+\sqrt{x-a})}\) =\(\displaystyle \lim_{x \to \infty} \frac{a\sqrt{x}}{(\sqrt{x}+\sqrt{x-a})}\) =\(\displaystyle \lim_{x \to \infty} \frac{a.1}{(1+\sqrt{1-\frac{a}{x}})}\) =\( \frac{a}{(1+\sqrt{1-0})}\) =\( \frac{a}{(1+1)}\) =\( \frac{a}{2}\)

7. Let \(f:R \to R \) be defined by \(f(x)=\begin{cases} x & \text{if x is an integer} \\ 0 & \text{if x is not an integer} \\ \end{cases}\)
   Find \( \displaystyle \lim_{x \to 1} f(x)\). Is it same as \(f(1)\)

   Solution 👉 Click Here

   Here

   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 1^-} x= \lim_{x \to 1^-} 0 = \) 0
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 1^+} x= \lim_{x \to 1^+} 0 = \) 0
   
   Since,
   LHL=RHL
   
   The limit exist at x=1, and the limit value is 0

   Next,
   The functional value at x=1, is
   \( f(x)= x =\) 1
   
   So, The functional value and limit value is NOT same.
8. Prove that
   1. \(\displaystyle \lim_{x \to 3} \left ( \frac{1}{x-3}-\frac{9}{x^3-3x^2}\right ) =\frac{2}{3}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 3} \left ( \frac{1}{x-3}-\frac{9}{x^3-3x^2}\right ) \)
      
      =\(\displaystyle \lim_{x \to 3} \left ( \frac{1}{x-3}-\frac{9}{x^2(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \frac{1}{x-3} \left ( 1-\frac{9}{x^2}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \frac{1}{x-3} \left ( \frac{x^2-9}{x^2}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x+3}{x^2}\right )\)
      
      = \(\frac{3+3}{3^2}\)
      
      = \(\frac{2}{3}\)
      
      

   2. \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9}{x^2-9}-\frac{3}{x-3}\right )=\frac{1}{2}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9}{x^2-9}-\frac{3}{x-3}\right )\)
      
      =\(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9}{(x+3)(x-3)}-\frac{3}{x-3}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2+9-3(x+3)}{(x+3)(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x^2-3x}{(x+3)(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x(x-3)}{(x+3)(x-3)}\right )\)
      
      = \(\displaystyle \lim_{x \to 3} \left ( \frac{x}{(x+3)}\right )\)
      
      = \(\frac{3}{3+3}\)
      
      = \(\frac{1}{2}\)
      
      

9. Evaluate
   1. \(\displaystyle \lim_{x \to 2} \frac{x^{-3}-2^{-3}}{x-2}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 2} \frac{x^{-3}-2^{-3}}{x-2}\)
      
      =\( (-3) (2)^{-3-1}\) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\), taking n=-3, a=2
      
      =\( (-3) (2)^{-4}\)
      
      =\( \frac{-3}{2^4}\)
      
      =\( \frac{-3}{16}\)
      
      

   2. \(\displaystyle \lim_{x \to \infty} \frac{(2x-1)^6(3x-1)^4}{(2x+1)^{10}}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to \infty} \frac{(2x-1)^6(3x-1)^4}{(2x+1)^{10}}\)
      
      =\(\displaystyle \lim_{x \to \infty} \frac{(2-\frac{1}{x})^6(3-\frac{1}{x})^4}{(2+\frac{1}{x})^{10}}\) Dividing by \(x^{10}\)
      
      =\(\frac{(2-0)^6(3-0)^4}{(2+0)^{10}}\)
      
      =\(\frac{3^4}{2^4}\)
      
      =\( \frac{81}{16}\)
      
      

   3. \(\displaystyle \lim_{x \to 0} \frac{(1+x)^6-1}{(1+x)^2-1}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to 0} \frac{(1+x)^6-1^6}{(1+x)^2-1^2}\)
      
      = \(\displaystyle \lim_{(1+x) \to 1} \frac{\frac{(1+x)^6-1^6}{(1+x)-1} }{\frac{(1+x)^2-1^2}{(1+x)-1} } \)
      
      =\( \frac{6.1^{6-1}}{2.1^{2-1}} \) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\)
      
      =\(\frac{6}{2}\)
      
      =\(3\)
      
      

   4. \(\displaystyle \lim_{x \to a} \frac{(x+2)^{\frac{5}{2}} -(a+2)^{\frac{5}{2}} }{x-a}\)

      Solution 👉 Click Here

      Solution
      
      \(\displaystyle \lim_{x \to a} \frac{(x+2)^{\frac{5}{2}} -(a+2)^{\frac{5}{2}} }{x-a}\)
      
      = \(\displaystyle \lim_{(x+2) \to (a+2)} \frac{(x+2)^{\frac{5}{2}} -(a+2)^{\frac{5}{2}} }{(x+2)-(a+2)}\)
      
      =\( \frac{5}{2}.(a+2)^{\frac{5}{2}-1} \) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\) taking \(n=\frac{5}{2}\), a=a+2
      
      =\( \frac{5}{2}.(a+2)^{\frac{3}{2}} \)
      
      

10. If \( \displaystyle \lim_{x \to a} \frac{x^3-a^3}{x-a}=27\), Find all possible values of a.

    Solution 👉 Click Here

    Solution
    
    \( \displaystyle \lim_{x \to a} \frac{x^3-a^3}{x-a}=27\)
    
    =\( 3.(a)^{3-1}=27 \) Using formula \(\displaystyle \lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}\) taking \(n=3\)
    
    =\( 3a^2=27 \)
    
    =\( a^2=9 \)
    
    =\( a=\pm 3 \)
    
    

11. Find the limiting values of
    1. \( \displaystyle \lim_{x \to 0} \frac{\sin x^0}{x}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to 0} \frac{\sin x^0}{x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin (\frac{\pi x}{180})}{x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin (\frac{\pi x}{180})}{\frac{\pi x}{180}} \frac{\pi}{180}\)
       
       =\( 1.\frac{\pi}{180} \)
       
       =\( \frac{\pi}{180} \)
       
       

    2. \( \displaystyle \lim_{x \to 0} \frac{1-\cos 4 x}{1-\cos 6x}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to 0} \frac{1-\cos 4 x}{1-\cos 6x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{1-\cos^2 4 x}{1+\cos 4 x}.\frac{1+\cos 6 x}{1-\cos^2 6x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin^2 4 x}{1+\cos 4 x}.\frac{1+\cos 6 x}{\sin^2 6x}\)
       
       = \( \displaystyle \lim_{x \to 0} \frac{\sin^2 4 x}{(4x)^2} . \frac{(6x)^2}{\sin^2 6 x} .\frac{1+\cos 6 x}{1+\cos 4 x} .\frac{(4x)^2}{(6x)^2}\)
       
       = \( \displaystyle \lim_{x \to 0} \left (\frac{\sin 4 x}{4x} \right )^2 \left (\frac{6x}{\sin 6 x} \right )^2 \left ( \frac{1+\cos 6 x}{1+\cos 4 x} \right ) \left (\frac{4^2}{6^2} \right ) \)
       
       = \( \left (1 \right )^2 \left (1 \right )^2 \left ( \frac{1+0}{1+0} \right ) \left (\frac{4}{9} \right ) \)
       
       = \( \frac{4}{9} \)
       
       

    3. \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2}-x}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2}-x}\)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin (\frac{\pi}{2}-x)} {(\frac{\pi}{2}-x)}\)
       
       = \(1\)
       
       

    4. \( \displaystyle \lim_{x \to 0} \frac{\tan 2 x-x}{3x-\sin x}\)

       Solution 👉 Click Here

       \( \displaystyle \lim_{x \to 0} \frac{\tan 2 x-x}{3x-\sin x}\)	=\( \displaystyle \lim_{x \to 0} \frac{\frac{\tan 2 x}{2}-1}{3-\frac{\sin x}{x}}\)
       	=\( \frac{\displaystyle \lim_{x \to 0} \frac{\tan 2 x}{2x}.2-1}{3-\displaystyle \lim_{x \to 0} \frac{\sin x}{x}}\) =\(\frac{1.2-1}{3-1}\) =\(\frac{1}{2}\)

    5. \( \displaystyle \lim_{x \to \pi} \frac{1-\sin (\frac{x}{2})}{(\pi-x)^2}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to \pi} \frac{1-\sin (\frac{x}{2})}{(\pi-x)^2}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{1-\cos (\frac{\pi}{2} -\frac{x}{2})}{(\pi-x)^2}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{1-\cos (\frac{\pi-x}{2})}{(\pi-x)^2}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{1-\cos^2 (\frac{\pi-x}{2})}{(\pi-x)^2} . \frac{1}{1+\cos (\frac{\pi-x}{2})}\)
       
       = \( \displaystyle \lim_{x \to \pi} \frac{\sin^2 (\frac{\pi-x}{2})}{(\pi-x)^2}. \frac{1}{1+\cos (\frac{\pi-x}{2})}\)
       
       = \( \displaystyle \lim_{x \to \pi} \left (\frac{\sin^2 (\frac{\pi-x}{2})}{(\frac{\pi-x}{2})} \right )^2. (\frac{1}{2})^2 .\frac{1}{1+\cos (\frac{\pi-x}{2})}\)
       
       = \( (1 )^2. \frac{1}{4} .\frac{1}{1+1}\)
       
       = \(\frac{1}{8}\)
       
       

    6. \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1+\cos 2x}{(\pi-2x)^2}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1+\cos 2x}{(\pi-2x)^2}\)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1-\cos (\pi-2x)}{(\pi-2x)^2}\)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1-\cos^2 (\pi-2x)}{(\pi-2x)^2}. \frac{1}{1+\cos (\pi-2x)} \)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin^2 (\pi-2x)}{(\pi-2x)^2}. \frac{1}{1+\cos (\pi-2x)} \)
       
       = \( \displaystyle \lim_{x \to \frac{\pi}{2}} \left ( \frac{\sin (\pi-2x)}{(\pi-2x)} \right )^2. \lim_{x \to \frac{\pi}{2}} \frac{1}{1+\cos (\pi-2x)} \)
       
       = \( (1)^2 \frac{1}{1+1} \)
       
       = \( \frac{1}{2} \)
       
       

    7. \( \displaystyle \lim_{x \to 0} \sin (\frac{1}{x})\)

       Solution 👉 Click Here

       Let \(\frac{1}{x}=u\) then
       
       \( \displaystyle \lim_{x \to 0} \sin (\frac{1}{x}) = \displaystyle \lim_{u \to \infty} \sin u \)
       
       The left hand limit is
       
       As \( x \to 0^-\) then \(\displaystyle \lim_{x \to 0^-} \frac{1}{x} =\lim_{h \to 0^-} \frac{1}{0-h}=-\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to -\infty} \sin (u)=\sin (-\infty)=- \sin \infty \)
       
       The right hand limit is
       
       As \( x \to 0^+\) then \(\displaystyle \lim_{x \to 0^+} \frac{1}{x} =\lim_{h \to 0^+} \frac{1}{0+h}=\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to \infty} \sin (u)= \sin (\infty) \)
       
       We see that, function \( \sin u\) goes up to infinity from the right. From the left,it goes down to negative infinity.
       So, the limit of a function \( \sin (\frac{1}{x}) \) does NOT exist at x=0.

       For more justification

       1. given the function \( f(x)=\sin \frac{1}{x} \)
       2. take \( x=\frac{1}{\frac{\pi}{2} (2n+1)};n \in Z \) then \( \frac{1}{x}=\frac{\pi}{2} (2n+1);n \in Z \) and \( \sin (\frac{1}{x})=\sin[\frac{\pi}{2} (2n+1)] \)
       3. then value of \(f(x)= \sin[\frac{\pi}{2} (2n+1)] \) are alternatively to 1 and -1 as x approaches to 0
       4. so, \( f(x)=\sin \frac{1}{x} \) has oscillating discontinuity at 0

       Figure 1

       Figure 2

       Figure 3

       Figure 4

       Figure 5
    8. \( \displaystyle \lim_{x \to 0} x \sin (\frac{1}{x})\)

       Solution 👉 Click Here

       Let \(\frac{1}{x}=u\) then
       
       \( \displaystyle \lim_{x \to 0} x \sin (\frac{1}{x}) = \displaystyle \lim_{u \to \infty} \frac{1}{u} \sin u \)
       
       The left hand limit is
       
       As \( x \to 0^-\) then \(\displaystyle \lim_{x \to 0^-} \frac{1}{x} =\lim_{h \to 0^-} \frac{1}{0-h}=-\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to -\infty} \frac{1}{u} \sin (u)= \frac{1}{\infty }\sin (-\infty)=0. (\sin \infty) =0\)
       
       The right hand limit is
       
       As \( x \to 0^+\) then \(\displaystyle \lim_{x \to 0^+} \frac{1}{x} =\lim_{h \to 0^+} \frac{1}{0+h}=\infty\)
       
       Thus
       
       \(\displaystyle \lim_{u \to \infty} \frac{1}{u} \sin (u)= \frac{1}{\infty} \sin (\infty) =0. \sin (\infty)=0\)
       
       We see that, function \( \frac{1}{u} \sin (u)\) goes up to 0 from both side.
       So, the limit of a function \( x \sin (\frac{1}{x}) \) exist at x=0, and the limit value is 0.

       For more justification
    9. \( \displaystyle \lim_{x \to a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}\)

       Solution 👉 Click Here

       Solution
       
       \( \displaystyle \lim_{x \to a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}\)
       
       = \( \displaystyle \lim_{x \to a} \frac{2 \cos (\frac{x+a}{2})\sin (\frac{x-a}{2})}{x-a}. (\sqrt{x}+\sqrt{a})\)
       
       = \( \left ( \displaystyle \lim_{x \to a} \frac{\sin (\frac{x-a}{2}) }{(\frac{x-a}{2}) } \right ) . \left ( \displaystyle \lim_{x \to a} \cos (\frac{x+a}{2}) .(\sqrt{x}+\sqrt{a}) \right )\)
       
       = \( (1) .\cos a .(2\sqrt{a}) \)
       
       = \( 2\sqrt{a}\cos a \)
       
       

    10. \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)

        Solution 👉 Click Here

        Solution
        \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{x \to a} (a-x)\cot (\frac{\pi }{2}-\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{x \to a} (a-x)\cot (\frac{\pi(a-x) }{2a})\)
        
        = \( \displaystyle \lim_{x \to a} \frac{a-x}{\tan (\frac{\pi(a-x) }{2a})}\)
        
        = \( \displaystyle \lim_{x \to a} \frac{(\frac{\pi(a-x) }{2a})} {\tan (\frac{\pi(a-x) }{2a}) } . (\frac{2a}{\pi}) \)
        
        = \( \frac{2a}{\pi}\)
        
        

        Solution Method 2

        Let a-x=h, then \(x \to a\) implies \(h \to 0\) and \( \frac{x}{a}=1+\frac{h}{a}\)

        Solution
        
        \( \displaystyle \lim_{x \to a} (a-x)\tan (\frac{\pi x}{2a})\)
        
        = \( \displaystyle \lim_{h \to 0} (h)\tan \frac{\pi}{2} (1+\frac{h}{a})\)
        
        = \( \displaystyle \lim_{h \to 0} (h)\tan (\frac{\pi}{2}+\frac{\pi h}{2 a})\)
        
        = \( \displaystyle \lim_{h \to 0} (h)\cot (\frac{\pi h}{2 a})\)
        
        = \( \displaystyle \lim_{h \to 0} \frac{h}{\tan (\frac{\pi h}{2 a})}\)
        = \( \displaystyle \lim_{h \to 0} \frac{(\frac{\pi h}{2 a})}{\tan (\frac{\pi h}{2 a})}. (\frac{2a}{\pi})\)
        
        = \( \frac{2a}{\pi}\)
        
        

    11. \( \displaystyle \lim_{y \to 0} \frac{(x+y)\sec (x+y)-x \sec x}{y}\)

        Solution 👉 Click Here

        \( \displaystyle \lim_{y \to 0} \frac{(x+y)\sec (x+y)-x \sec x}{y}\)

        =\( \displaystyle \lim_{y \to 0} \frac{x\sec (x+y)+y\sec (x+y)-x \sec x}{y}\)

        =\( \displaystyle \lim_{y \to 0} \frac{x\sec (x+y)-x \sec x}{y} + \lim_{y \to 0} \frac{y\sec (x+y)}{y} \) =\( \displaystyle x \lim_{y \to 0} \frac{\sec (x+y)- \sec x}{y} +\lim_{y \to 0} \sec (x+y) \) =\( \displaystyle x \lim_{y \to 0} \frac{\cos x-\cos (x+y)}{y\cos x \cos (x+y)} +\sec x \) =\( \displaystyle x \lim_{y \to 0} \frac{-2 \sin (\frac{x+x+y}{2}) \sin (\frac{x-x-y}{2})}{y\cos x \cos (x+y)} +\sec x \) =\( \displaystyle x \lim_{y \to 0} \frac{2 \sin (x+\frac{y}{2}) \sin (\frac{y}{2})}{y\cos x \cos (x+y)} +\sec x \) =\( \displaystyle x \lim_{y \to 0} \frac{ \sin (x+\frac{y}{2}) }{\cos x \cos (x+y)}.\lim_{y \to 0} \frac{ \sin (\frac{y}{2})}{\frac{y}{2}} +\sec x \) =\( x \frac{ \sin (x+0) }{\cos x \cos (x+0)}.1 +\sec x \) =\( x \frac{ \sin x }{\cos x \cos x} +\sec x \) =\( x \tan x \sec x +\sec x \)

12. 
    
    1. A function is defined as \(f(x)=\begin{cases} 3x^2+2 & \text { if } x<1 \\ 2x+3 & \text { if } x \ge 1 \end{cases}\).
       Find \(\displaystyle \lim_{x \to 1}f(x) \)

       Solution 👉 Click Here

       Solution
       
       

       The left hand limit is
       \( LHL=\displaystyle \lim_{x \to 1^-} 3x^2+2= 3.1^2+2= \) 5
       
       The right hand limit is
       \(RHL= \displaystyle \lim_{x \to 1^+} 2x+3= 2.1+3= \) 5
       
       Since,
       LHL=RHL
       
       The limit exist at x=1, and the limit value is 5
    2. A function is defined as \(f(x)=\begin{cases} 3+2x & \text { if } -\frac{3}{2} \le x < 0\\ 3-2x & \text { if } 0 \le x < \frac{3}{2} \\ -3-2x & \text { if } x \ge \frac{3}{2} \end{cases}\).
       Find \(\displaystyle \lim_{x \to 0}f(x) \) and \(\displaystyle \lim_{x \to \frac{3}{2}}f(x) \) if they exist.

       Solution 👉 Click Here

       Solution \(\displaystyle \lim_{x \to 0}f(x) \)
       
       

       The left hand limit is
       \( LHL=\displaystyle \lim_{x \to 0^-} 3+2x= 3+2.0 = \) 3
       
       The right hand limit is
       \(RHL= \displaystyle \lim_{x \to 0^+} 3-2x=3-2.0 = \) 3
       
       Since,
       LHL=RHL
       
       The limit exist at x=0, and the limit value is 3

       Solution \(\displaystyle \lim_{x \to 3/2}f(x) \)
       
       

       The left hand limit is
       \( LHL=\displaystyle \lim_{x \to \frac{3}{2}^-} 3-2x= 3-3= \) 0
       
       The right hand limit is
       \(RHL= \displaystyle \lim_{x \to \frac{3}{2}^+} -3-2x=-3-3 = \) -6
       
       Since,
       LHL≠RHL
       
       The limit does NOT exist at x=3/2

13. 
    
    1. \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{e^{qx}-1} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{e^{qx}-1} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{px} . \frac{qx}{e^{qx}-1} . \frac{px}{qx}\)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{px}-1}{px} . \lim_{x \to 0} \frac{qx}{e^{qx}-1} . \frac{p}{q}\)
       
       = \(1.1 . \frac{p}{q}\) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}=1\)
       
       = \(\frac{p}{q}\)
       
       

    2. \(\displaystyle \lim_{x \to 0} \frac{e^x-e^{-x}-x}{x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{e^x-e^{-x}-x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^x-1-e^{-x}+1-x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}-\frac{e^{-x}-1}{x}-\frac{x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}+\lim_{x \to 0} \frac{e^{-x}-1}{-x}-1\)
       
       = \(1+1-1\) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x}=1\)
       
       = \(1+1-1\)
       
       

    3. \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{b^x-1} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{b^x-1} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{x} . \frac{x}{b^x-1} \) Using formula \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{x}=\log a\)
       
       = \(\log a. \frac{1}{\log b}\)
       
       = \(\log (a-b)\)
       
       

14. 
    
    1. \(\displaystyle \lim_{x \to 0} \frac{2^x-1}{\sin x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{2^x-1}{\sin x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{2^x-1}{x}. \lim_{x \to 0} \frac{x}{\sin x} \) Using formula \(\displaystyle \lim_{x \to 0} \frac{a^x-1}{x}=\log a\), taking a=2
       
       = \(\log 2. 1\)
       
       = \(\log 2\)
       
       

    2. \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x} -\sin x -1}{x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x} -\sin x -1}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x}-1}{x} -\lim_{x \to 0} \frac{\sin x}{x} \)
       
       = \(\displaystyle \lim_{x \to 0} \frac{e^{\sin x}-1}{\sin x} .\lim_{x \to 0} \frac{\sin x}{x} - \lim_{x \to 0} \frac{\sin x}{x} \)
       
       = \( 1 .1 - 1 \) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} =1\)
       
       = \(0\)
       
       

    3. \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\cos x} -1}{\frac{\pi}{2}-x} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\cos x} -1}{\frac{\pi}{2}-x} \)
       
       = \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\sin (\frac{\pi}{2}-x)} -1}{\frac{\pi}{2}-x} \)
       
       = \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{e^{\sin (\frac{\pi}{2}-x)} -1}{\sin (\frac{\pi}{2}-x)} . \lim_{x \to 0} \frac{\sin (\frac{\pi}{2}-x)}{\frac{\pi}{2}-x} \)
       
       = \(1 .1 \) Using formula \(\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} =1\)
       
       = \(1 \)
       
       

    4. \(\displaystyle \lim_{x \to e} \frac{\log x-1}{x-e} \)

       Solution 👉 Click Here

       Solution
       
       \(\displaystyle \lim_{x \to e} \frac{\log x-1}{x-e} \)
       
       = \(\displaystyle \lim_{x \to e} \frac{1}{e}.\frac{\log x-\log e}{\frac{x}{e}-1} \)
       
       = \(\frac{1}{e} . \displaystyle \lim_{x \to e} \frac{\log (\frac{x}{e})}{\frac{x}{e}-1} \)
       
       = \(\frac{1}{e} . \displaystyle \lim_{x \to e} \frac{\log (1+[\frac{x}{e}-1])}{[\frac{x}{e}-1]} \)
       
       = \(\frac{1}{e} \) Using formula \(\displaystyle \lim_{x \to 0} \frac{\log (1+x)}{x} =1\)