If f(x) = begin{cases} frac{x^2 - 1}{x + 1}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = \begin{cases} \frac{x^2 - 1}{x + 1}, & x 
eq -1 \ -2, & x = -1 \end{cases}$.For $x 
eq -1$,$f(x) = \frac{(x - 1)(x + 1)}{x + 1} =

Vedclass · Accepted Answer

All the above are correct

Vedclass · Answer

(D) Given $f(x) = \begin{cases} \frac{x^2 - 1}{x + 1}, & x 
eq -1 \ -2, & x = -1 \end{cases}$.For $x 
eq -1$,$f(x) = \frac{(x - 1)(x + 1)}{x + 1} = x - 1$.Now,calculate the left-hand limit: $\lim_{x 	o (-1)^-} f(x) = \lim_{x 	o -1} (x - 1) = -1 - 1 = -2$.Calculate the right-hand limit: $\lim_{x 	o (-1)^+} f(x) = \lim_{x 	o -1} (x - 1) = -1 - 1 = -2$.Since $\lim_{x 	o (-1)^-} f(x) = \lim_{x 	o (-1)^+} f(x) = f(-1) = -2$,the function is continuous at $x = -1$.Therefore,all the given statements are correct.

If $f(x) = \begin{cases} \frac{x^2 - 1}{x + 1}, & x \neq -1 \\ -2, & x = -1 \end{cases}$,then which of the following is true?

Similar Questions

If $f(x) = \begin{cases} \frac{x}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then

If the function $f(x) = \begin{cases} \frac{\cos ax - \cos 9x}{x^2}, & x \neq 0 \\ 16, & x = 0 \end{cases}$ is continuous at $x = 0$, then $a =$

The number of discontinuities in $\mathbb{R}$ for the function $f(x)=\frac{x-1}{x^3+6x^2+11x+6}$ is

Find all points of discontinuity of $f,$ where $f$ is defined by $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$

The value of $f(0)$,so that the function $f(x) = \frac{(27 - 2x)^{1/3} - 3}{9 - 3(243 + 5x)^{1/5}}, (x \ne 0)$ is continuous,is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module