Contunuity of a function [Extra Exercise]

1. Test the continuity of a function f(x)=x²+1 at x=0

   Solution 👉 Click Here

   Given that
   \( f(x)=x^2+1 \)
   
   At x=0, we compute the following
   The functional value is
   \(f(x)= x^2+1 = 0^2+1=\) 1
   
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0^-} x^2+1 = 0^2+1=\) 1
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0^+} x^2+1 = 0^2+1=\) 1
   
   Since,
   LHL=RHL=f(x)
   
   The function is continuous at x=0

   
   This completes the solution
2. Test the continuity of a function \( f(x)= \begin{cases} \frac{|x|}{x} & x \ne 2\\ x & x =2 \end{cases} \) at x=2.

   Solution 👉 Click Here

   Given that
   \( f(x)= \begin{cases} \frac{|x|}{x} & x \ne 2\\ x & x =2 \end{cases} \)
   
   At x=2, we compute the following
   The functional value is
   \( f(x)= x= \) 2
   
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 2^-} \frac{|x|}{x}= \lim_{x \to 2^-} \frac{x}{x} = \) 1
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 2^+} \frac{|x|}{x}= \lim_{x \to 2^+} \frac{x}{x} = \) 1
   
   Since,
   LHL=RHL≠f(x)
   
   The function is NOT continuous at x=2

   
   This completes the solution
3. Test the continuity of a function \( f(x)= \begin{cases} 4x-3 & x < 2\\ (x-3)^2 & x ≥ 2 \end{cases} \) at x=2.

   Solution 👉 Click Here

   Given that
   \( f(x)= \begin{cases} 4x-3 & x < 2\\ (x-3)^2 & x ≥ 2 \end{cases} \)
   
   At x=2, we compute the following
   The functional value is
   \(f(x)= (x-3)^2= (2-3)^2=\) 1
   
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} 4x-3 = 4.2-3=\) 5
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}(x-3)^2= (2-3)^2=\) 1
   
   Since,
   LHL≠RHL=f(x)
   
   The function is NOT continuous at x=2

   
   This completes the solution
4. Test the continuity of a function \( f(x)= \begin{cases} \frac{x^2-4}{x-2} & x \ne 2\\ 4 & x =2 \end{cases} \) at x=2.

   Solution 👉 Click Here

   Given that
   \( f(x)= \begin{cases} \frac{x^2-4}{x-2} & x \ne 2\\ 4 & x =2 \end{cases} \)
   
   At x=2, we compute the following
   The functional value is
   \(f(x)= \) 4
   
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2^-} \frac{x^2-4}{x-2}=x+2=2+2=\) 4
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2^+}\frac{x^2-4}{x-2}=x+2=2+2=\) 4
   
   Since,
   LHL=RHL=f(x)
   
   The function is continuous at x=2

   
   This completes the solution
5. Test the continuity of a function \(f(x)=\begin{cases} \frac{1}{x} & x < -1 \\ \frac{x-1}{2} & -1 \le x < 1 \\ \sqrt{x} & x \ge 1 \\ \end{cases} \) at x=-1.

   Solution 👉 Click Here

   Given that
   \(f(x)=\begin{cases} \frac{1}{x} & x < -1 \\ \frac{x-1}{2} & -1 \le x < 1 \\ \sqrt{x} & x \ge 1 \\ \end{cases} \)
   
   At x=-1, we compute the following
   The functional value is
   \(f(x)=\frac{x-1}{2}=\frac{-1-1}{2}=\) -1
   
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to -1^-} f(x)= \lim_{x \to -1^-} \frac{x-1}{2}=\frac{-1-1}{2}=\) -1
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to -1^+} f(x)= \lim_{x \to -1^+}\frac{x-1}{2}=\frac{-1-1}{2}=\) -1
   
   Since,
   LHL=RHL=f(x)
   
   The function is continuous at x=-1

   
   This completes the solution
6. Test the continuity of a function \( f(x)=\begin{cases} x-2 & x < 1 \\ \sqrt{x} & x \ge 1 \\ \end{cases} \) at x=-1 and at x=1.

   Solution 👉 Click Here

   Testing the continuity of the function at x=-1

   Given that
   \( f(x)=\begin{cases} x-2 & x < 1 \\ \sqrt{x} & x \ge 1 \\ \end{cases} \)
   
   At x=-1, we compute the following
   The functional value is
   \(f(x)=x-2=-1-2=\) -3
   
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to -1^-} f(x)= \lim_{x \to -1^-} x-2=-1-2=\) -3
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to -1^+} f(x)= \lim_{x \to -1^+}x-2=-1-2=\) -3
   
   Since,
   LHL=RHL=f(x)
   
   The function is continuous at x=-1

   
   This completes the solution

   Testing the continuity of the function at x=1

   Given that
   \( f(x)=\begin{cases} x-2 & x < 1 \\ \sqrt{x} & x \ge 1 \\ \end{cases} \)
   
   At x=1, we compute the following
   The functional value is
   \(f(x)=\sqrt{x}=\sqrt{1}=\) 1
   
   The left hand limit is
   \( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1^-} x-2=1-2=\) -1
   
   The right hand limit is
   \(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1^+}\sqrt{x}=\sqrt{1}=\) 1
   
   Since,
   LHL≠RHL=f(x)
   
   The function is NOT continuous at x=1

   
   This completes the solution
7. For what value of k would make the function continuous in each case? Explain how you found your value.
   
   • \( f(x)= \begin{cases} \frac{x^2-9}{x+3} & x\ne -3 \\ k & x=-3 \end{cases} \) at x=-3
     
     

     Solution 👉 Click Here

     Given that
     \( f(x)= \begin{cases} \frac{x^2-9}{x+3} & x\ne -3 \\ k & x=-3 \end{cases} \)
     
     At x=-3, we compute the following
     The functional value is
     \(f(x)=k=\) k
     
     The left hand limit is
     \( LHL=\displaystyle \lim_{x \to -3^-} f(x)= \lim_{x \to -3^-} \frac{x^2-9}{x+3} =x-3=-3-3=\) -6
     
     The right hand limit is
     \(RHL= \displaystyle \lim_{x \to -3^+} f(x)= \lim_{x \to -3^+}\frac{x^2-9}{x+3} =x-3=-3-3=\) -6
     
     Since,
     Function is assumed continuous at x=-3, thus we must have
     LHL=RHL=f(x)
     
     or k=-6
     

     
     This completes the solution
   • \( f(x)= \begin{cases} \frac{\sin (5\pi x)-1}{2x-1} & x\ne \frac{1}{2} \\ k & x= \frac{1}{2} \end{cases} \) at \( x=\frac{1}{2}\)
     
     

     Solution 👉 Click Here

     Given that
     \( f(x)= \begin{cases} \frac{\sin (5\pi x)-1}{2x-1} & x\ne \frac{1}{2} \\ k & x= \frac{1}{2} \end{cases} \)
     
     At x=1/2, we compute the following
     The functional value is
     \(f(x)=k=\) k
     
     The left hand limit is
     

     LHL

     =\(\displaystyle \lim_{x \to \frac{1}{2}^-} f(x)\)

     =\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{\sin (5 \pi x)-1}{2x-1} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{\sin (\pi x)-1 }{2x-1} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{\sin (\pi x)-\sin (\frac{\pi}{2}) }{2x-1} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{2 \cos \left ( \frac{\pi x+\frac{\pi}{2}}{2}\right ). \sin \left ( \frac{\pi x+\frac{\pi}{2}}{2}\right ) }{2x-1} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{2 \cos \frac{\pi}{4} (2x+1). \sin \frac{\pi}{4} (2x-1)}{2x-1} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-}\frac{2 \cos \frac{\pi}{4} (2x+1)}.\frac{\sin \frac{\pi}{4} (2x-1)}{2x-1} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-} 2 \cos \left ( \frac{\pi}{4} (2x+1) \right ).\frac{\sin \frac{\pi}{4} (2x-1)}{\frac{\pi}{4}(2x-1)}. \frac{\pi}{4} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-}2 \cos \left ( \frac{\pi}{4} (2x+1) \right ). \displaystyle \lim_{x \to \frac{1}{2}^-} \frac{\sin \frac{\pi}{4} (2x-1)}{\frac{\pi}{2}(2x-1)}. \frac{\pi}{2} \) =\(\displaystyle \lim_{x \to \frac{1}{2}^-} 2 \cos \left ( \frac{\pi}{4} (2x+1) \right ). \frac{\pi}{2} \) =\(2 \cos (\frac{\pi}{4} .2). \frac{\pi}{4} \) =\(2 \cos (\frac{\pi}{2}). \frac{\pi}{4} \) =\(2 .0. \frac{\pi}{4} \) =0

     Similarly, the right hand limit is
     \(RHL= \displaystyle \lim_{x \to \frac{1}{2}^+} f(x)=\) 0
     
     Since,
     Function is assumed continuous at x=1/2, thus we must have
     LHL=RHL=f(x)
     
     or k=0
     

     
     This completes the solution
   • \( f(x)= \begin{cases} \frac{e^x-1}{x} & x\ne 0 \\ k & x=0 \end{cases} \) at x=0
     

     Solution 👉 Click Here

     Given that
     \( f(x)= \begin{cases} \frac{e^x-1}{x} & x\ne 0 \\ k & x=0 \end{cases} \)
     
     At x=0, we compute the following
     The functional value is
     \(f(x)=k=\) k
     
     The left hand limit is
     \(\displaystyle \lim_{x\to 0^-}\frac{e^x-1 }{x} \)
     
     or \(\displaystyle \lim_{x\to 0}\frac{\left ( 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right ) -1 }{x} \)
     
     or \(\displaystyle \lim_{x\to 0}\frac{\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+... }{x} \)
     
     or \(\displaystyle \lim_{x\to 0} \frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+... \)
     
     or \(\frac{1}{1!}+\frac{0}{2!}+\frac{0^2}{3!}+... \)
     
     or 1
     
     Similarly, the left hand limit is
     \(\displaystyle \lim_{x\to 0^-}\frac{e^x-1 }{x} \)
     
     or 1
     
     Since,
     Function is assumed continuous at x=0, thus we must have
     LHL=RHL=f(x)
     
     or k=1
     

     
     This completes the solution.

8. Evaluate \( \displaystyle \lim_{x \to 4} \frac{(x-3)}{(4-x)(x+3)}\)
   Solution
   

   Solution 👉 Click Here

   Given that
   \( \displaystyle \lim_{x \to 4} \frac{(x-3)}{(4-x)(x+3)}\)
   
   At x=4, we compute the following
   The functional value is
   \(f(x)=\displaystyle \lim_{x \to 4} \frac{(x-3)}{(4-x)(x+3)}=\) Not Defined
   
   The left hand limit is
   

   LHL

   =\(\displaystyle \lim_{x \to 4^-} f(x)\)

   =\(\displaystyle \lim_{x \to 4^-} \frac{(x-3)}{(4-x)(x+3)}\) =\(\displaystyle \lim_{x \to 4} \frac{1}{4-x } \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) =\(\displaystyle \lim_{h \to 0} \frac{1}{4-(4-h)} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) =\(\displaystyle \lim_{hx \to 0} \frac{1}{4-4+h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) =\(\displaystyle \lim_{hx \to 0} \frac{1}{h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) = \(\infty\)

   The left hand limit is
   

   LHL

   =\(\displaystyle \lim_{x \to 4^+} f(x)\)

   =\(\displaystyle \lim_{x \to 4^+} \frac{(x-3)}{(4-x)(x+3)}\) =\(\displaystyle \lim_{x \to 4^+} \frac{1}{4-x } \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) =\(\displaystyle \lim_{h \to 0} \frac{1}{4-(4+h)} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) =\(\displaystyle \lim_{hx \to 0} \frac{1}{4-4-h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) =\(\displaystyle \lim_{hx \to 0} \frac{1}{-h} \displaystyle \lim_{x \to 4} \frac{(x-3)}{(x+3)}\) = \(- \infty\)

   Since,
   LHL≠RHL, and f(x) is NOT defined
   
   limit does NOT exist. Function is NOT continuous at x=4

   
   This completes the solution

   Method 2

   If we substitute 𝑥=4 in the function \( \frac{(x-3)}{(4-x)(x+3)}\) then we get \(\frac{1}{0}\), so we do as follows.

   	(-∞,-3)	(-3,3)	(3,4)	(4,∞)
   (x-3)	-	-	+	+
   (4-x)	+	+	+	-
   (x+3)	-	+	+	+
   Result			+	-

   So, LHS=+∞, and RHS=-∞, therefor limit does NOT exist.