Limit (BCB-Revised Edition 2020, Exercise 15.3, Page 389)

By Bed Prasad Dhakal Continuity · Grade 11

Test the continuity of discontinuity of the following function by calculating the left hand limits, the right-hand limits and the values of the functions at points specified.
1. \( f(x)=x^2\) at x=4
  Solution 👉 Click Here
  
  Given that
  \( f(x)=x^2 \)
  
  At x=4, we compute the following
  The functional value is
  \(f(x)= x^2 = 4^2=\) 16
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 4^-} f(x)= \lim_{x \to 4^-} x^2 = 4^2=\) 16
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 4^+} f(x)= \lim_{x \to 4^+} x^2 = 4^2=\) 16
  
  Since,
  LHL=RHL=f(x)
  
  The function is continuous at x=4
  
  This completes the solution
2. \( f(x)=2-3x^2\) at x=0
  Solution 👉 Click Here
  
  Given that
  \( f(x)=2-3x^2 \)
  
  At x=0, we compute the following
  The functional value is
  \(f(x)= 2-3x^2= 2-3.0^2=\) 2
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0} 2-3x^2= 2-3.0^2=\) 2
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0} 2-3x^2= 2-3.0^2=\) 2
  
  Since,
  LHL=RHL=f(x)
  
  The function is continuous at x=0
  
  This completes the solution
3. \( f(x)=3x^2-2x+4\) at x=1
  Solution 👉 Click Here
  
  Given that
  \( f(x)=3x^2-2x+4 \)
  
  At x=1, we compute the following
  The functional value is
  \(f(x)= 3x^2-2x+4=3.1^2-2.1+4=\) 5
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1} 3x^2-2x+4=3.1^2-2.1+4=\) 5
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1} 3x^2-2x+4=3.1^2-2.1+4=\) 5
  
  Since,
  LHL=RHL=f(x)
  
  The function is continuous at x=1
  
  This completes the solution
4. \( f(x)=\frac{1}{2x}\) at x=0
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{1}{2x} \)
  
  At x=0, we compute the following
  The functional value is
  \(f(x)= \frac{1}{2x}=\frac{1}{2.(0)}=\) ∞
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{h \to 0^-} \frac{1}{2(0-h)}= \lim_{h \to 0} -\frac{1}{2h}=\) -∞
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{h \to 0^+} \frac{1}{2(0+h)}= \lim_{h \to 0} \frac{1}{2h}=\) ∞
  
  Since,
  LHL,RHL,f(x) are NOT equal
  
  The function is NOT continuous at x=0
  
  This completes the solution
5. \( f(x)=\frac{1}{x-2}\) at x≠2
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{1}{x-2} \)
  
  At x=2, we compute the following
  The functional value is
  \(f(x)= \frac{1}{x-2}=\) NOT defined
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{h \to 0^-} \frac{1}{(2-h)-2}= \lim_{h \to 0} \frac{1}{-h}=\) -∞
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{h \to 0^+} \frac{1}{(2+h)-2}= \lim_{h \to 0} \frac{1}{h}=\) ∞
  
  Since,
  LHL ≠RHL,f(x) is NOT defined
  
  The function is NOT continuous at x=2
  
  This completes the solution
6. \( f(x)=\frac{1}{3x}\) at x≠0
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{1}{3x} \)
  
  At x=0, we compute the following
  The functional value is
  \(f(x)= \frac{1}{3x}=\) NOT defined
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{h \to 0^-} \frac{1}{3(0-h)}= \lim_{h \to 0} -\frac{1}{3h}=\) -∞
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{h \to 0^+} \frac{1}{3(0+h)}= \lim_{h \to 0} \frac{1}{3h}=\) ∞
  
  Since,
  LHL ≠RHL,f(x) is NOT defined
  
  The function is NOT continuous at x=0
  
  This completes the solution
7. \( f(x)=\frac{1}{1-x}\) at x=1
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{1}{1-x} \)
  
  At x=1, we compute the following
  The functional value is
  \(f(x)= \frac{1}{1-x}=\) ∞
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{h \to 0^-} \frac{1}{1-(1-h)}= \lim_{h \to 0} \frac{1}{h}=\) ∞
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{h \to 0^+} \frac{1}{1-(1+h)}= \lim_{h \to 0} \frac{1}{-h}=\) -∞
  
  Since,
  LHL ,RHL,f(x) are NOT equal
  
  The function is NOT continuous at x=1
  
  This completes the solution
8. \( f(x)=\frac{1}{x-3}\) at x=3
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{1}{x-3} \)
  
  At x=3, we compute the following
  The functional value is
  \(f(x)= \frac{1}{x-3}=\) ∞
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{h \to 0^-} \frac{1}{(3-h)-3}= \lim_{h \to 0} \frac{1}{-h}=\) -∞
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{h \to 0^+} \frac{1}{(3+h)-3}= \lim_{h \to 0} \frac{1}{h}=\) ∞
  
  Since,
  LHL ≠RHL,f(x) are NOT equal
  
  The function is NOT continuous at x=3
  
  This completes the solution
9. \( f(x)=\frac{x^2-9}{x-3}\) at x=3
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{x^2-9}{x-3}\)
  
  At x=3, we compute the following
  The functional value is
  \(f(x)= \frac{x^2-9}{x-3}=x+3=3+3=\) 6
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3^-} \frac{x^2-9}{x-3}=\lim_{x \to 3}x+3=3+3=\) 6
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3^+} \frac{x^2-9}{x-3}=\lim_{x \to 3}x+3=3+3=\) 6
  
  Since,
  LHL=RHL=f(x)
  
  The function is continuous at x=3
  
  This completes the solution
10. \( f(x)=\frac{x^2-16}{x-4}\) at x=4
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{x^2-16}{x-4}\)
  
  At x=4, we compute the following
  The functional value is
  \(f(x)= \frac{x^2-16}{x-4}=x+4=4+4=\) 8
  
  The left hand limit is
  \( LHL=\displaystyle \lim_{x \to 4^-} f(x)= \lim_{x \to 4^-} \frac{x^2-16}{x-4}=\lim_{x \to 4}x+4=4+4=\) 8
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 4^+} f(x)= \lim_{x \to 4^+} \frac{x^2-9}{x-3}=\lim_{x \to 4}x+4=4+4=\) 8
  
  Since,
  LHL=RHL=f(x)
  
  The function is continuous at x=4
  
  This completes the solution
11. \( f(x)=\frac{|x-2|}{x-2}\) at x=2
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{|x-2|}{x-2}\)
  
  At x=2, we compute the following
  The functional value is
  \(f(x)=\frac{|x-2|}{x-2}=\) NOT defined
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} \frac{-(x-2)}{x-2} = \) -1
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} \frac{(x-2)}{x-2}=\) 1
  
  Since,
  LHL≠RHL, f(x) is NOT defined
  
  The function is NOT continuous at x=2
  
  This completes the solution
12. \( f(x)=\frac{x}{|x|}\) at x=0
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\frac{x}{|x|}\)
  
  At x=0, we compute the following
  The functional value is
  \(f(x)=\frac{x}{|x|}=\) NOT defined
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 0^-} f(x)= \lim_{x \to 0} \frac{x}{-x} = \) -1
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 0^+} f(x)= \lim_{x \to 0} \frac{x}{x}=\) 1
  
  Since,
  LHL≠RHL, f(x) is NOT defined
  
  The function is NOT continuous at x=2
  
  This completes the solution
Discuss the continuity of functions at the points specified
1. \( f(x)= \begin{cases} 2-x^2 & \text{for } x \le 2 \\ 1 & \text{for } x > 2 \end{cases} \) at x=2
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} 2-x^2 & \text{for } x \le 2 \\ 1 & \text{for } x > 2 \end{cases}\)
  
  At x=2, we compute the following
  The functional value is
  \(f(x)=2-x^2=2-2^2=\) -2
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 2-x^2=2-2^2=\) -2
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} 1=\) 1
  
  Since,
  LHL,RHL, f(x) are NOT equal
  
  The function is NOT continuous at x=2
  
  This completes the solution
2. \( f(x)= \begin{cases} 2x^2+4 & \text{for } x \le 2 \\ 4x+1 & \text{for } x > 2 \end{cases} \) at x=2
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} 2x^2+4 & \text{for } x \le 2 \\ 4x+1 & \text{for } x > 2 \end{cases}\)
  
  At x=2, we compute the following
  The functional value is
  \(f(x)=2x^2+4=2.2^2+4=\) 12
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 2x^2+4=2.2^2+4=\) 12
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} 4x+1=4.2+1=\) 9
  
  Since,
  LHL,RHL, f(x) are NOT equal
  
  The function is NOT continuous at x=2
  
  This completes the solution
3. \( f(x)= \begin{cases} 2x & \text{for } x \le 3 \\ 3x-3 & \text{for } x > 3 \end{cases} \) at x=3
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} 2x & \text{for } x \le 3 \\ 3x-3 & \text{for } x > 3 \end{cases}\)
  
  At x=3, we compute the following
  The functional value is
  \(f(x)=2x=2.3=\) 6
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3} 2x=2.3=\) 6
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3} 3x-3=3.3-3=\) 6
  
  Since,
  LHL,RHL, f(x) are equal
  
  The function is continuous at x=3
  
  This completes the solution
4. \( f(x)= \begin{cases} 2x+1 & \text{for } x < 1 \\ 2 & \text{for } x =1 \\ 3x & \text{for } x >1 \end{cases} \) at x=1
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} 2x+1 & \text{for } x < 1 \\ 2 & \text{for } x =1 \\ 3x & \text{for } x >1 \end{cases}\)
  
  At x=1, we compute the following
  The functional value is
  \(f(x)=\) 2
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 1^-} f(x)= \lim_{x \to 1} 2x+1=2.1+1=\) 3
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 1^+} f(x)= \lim_{x \to 1} 3x=3.1=\) 3
  
  Since,
  LHL,RHL, f(x) are NOT equal
  
  The function is NOT continuous at x=1
  
  This completes the solution
1. A function is defined as follows
  \( f(x)= \begin{cases} x^2+2 & \text{for } x < 5 \\ 20 & \text{for } x =5 \\ 3x+12 & \text{for } x > 5 \end{cases} \).
  Show that f(x) has removable discontinuity at x=5
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} x^2+2 & \text{for } x < 5 \\ 20 & \text{for } x =5 \\ 3x+12 & \text{for } x > 5 \end{cases} \)
  
  At x=5, we compute the following
  The functional value is
  \(f(x)=\) 20
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 5^-} f(x)= \lim_{x \to 5} x^2+2=5^2+2=\) 27
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 5^+} f(x)= \lim_{x \to 5} 3x+12=3.5+12=\) 27
  
  Since,
  LHL=RHL≠f(x)
  
  The function has removable discontinuity at x=5
  
  This completes the solution
2. A function is defined as follows
  \( f(x)= \begin{cases} 2x-3 & \text{for } x < 2 \\ 2 & \text{for } x =2 \\ 3x-5 & \text{for } x > 2 \end{cases} \).
  Is the function f(x) continuous at x=2? If not, how can the function f(x) be made continuous at x=2?
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} 2x-3 & \text{for } x < 2 \\ 2 & \text{for } x =2 \\ 3x-5 & \text{for } x > 2 \end{cases} \)
  
  At x=2, we compute the following
  The functional value is
  \(f(x)=\) 2
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 2x-3=2.2-3=\) 1
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} 3x-5=3.2-5=\) 1
  
  Since,
  LHL=RHL≠f(x)
  
  The function has removable discontinuity at x=2, so the function f(x) be made continuous at x=2 be redefining the function as below.
  \( f(x)= \begin{cases} 2x-3 & \text{for } x < 2 \\ 1 & \text{for } x =2 \\ 3x-5 & \text{for } x > 2 \end{cases} \).
  This completes the solution
1. A function is defined as follows
  \( f(x)= \begin{cases} kx+3 & \text{for } x \ge 2 \\ 3x-1 & \text{for } x < 2 \end{cases} \).
  Find the value of k so that f(x) is continuous at x=2
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} kx+3 & \text{for } x \ge 2 \\ 3x-1 & \text{for } x < 2 \end{cases} \)
  
  At x=2, we compute the following
  The functional value is
  \(f(x)=kx+3=\) 2k+3
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 2^-} f(x)= \lim_{x \to 2} 3x-1=3.2-1=\) 5
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 2^+} f(x)= \lim_{x \to 2} kx+3=\) 2k+3
  
  Since the functinuous at x=2, we must have
  LHL=RHL=f(x)
  
  or 2k+3=5
  
  or k=1
  
  This completes the solution
2. A function is defined as follows
  \( f(x)= \begin{cases} \frac{2x^2-18}{x-3} & \text{for } x \ne 3 \\ k & \text{for } x =3 \end{cases} \).
  Find the value of k so that f(x) is continuous at x=3
  Solution 👉 Click Here
  
  Given that
  \( f(x)=\begin{cases} \frac{2x^2-18}{x-3} & \text{for } x \ne 3 \\ k & \text{for } x =3 \end{cases} \)
  
  At x=3, we compute the following
  The functional value is
  \(f(x)=\) k
  
  The left hand limit is
  \(LHL=\displaystyle \lim_{x \to 3^-} f(x)= \lim_{x \to 3} \frac{2x^2-18}{x-3}=2{x+3}=2{3+3}=\) 12
  
  The right hand limit is
  \(RHL= \displaystyle \lim_{x \to 3^+} f(x)= \lim_{x \to 3} \frac{2x^2-18}{x-3}=2{x+3}=2{3+3}=\) 12
  
  Since the functinuous at x=3, we must have
  LHL=RHL=f(x)
  
  or k=12
  
  This completes the solution

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