mathop {lim }limits_{x to 1} f(x) is equal to | vedclass

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

Question

(D) To find $\mathop {\lim }\limits_{x 	o 1} f(x)$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.$LHL = \mathop {\lim }\limits

Vedclass · Accepted Answer

does not exist

Vedclass · Answer

(D) To find $\mathop {\lim }\limits_{x 	o 1} f(x)$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.$LHL = \mathop {\lim }\limits_{x 	o 1^-} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2}$. Let $x = 1 - h$,where $h 	o 0^+$.$LHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{e^{-\frac{1}{h}} - 2}{e^{-\frac{1}{h}} + 2} = \frac{0 - 2}{0 + 2} = -1$.$RHL = \mathop {\lim }\limits_{x 	o 1^+} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2}$. Let $x = 1 + h$,where $h 	o 0^+$.$RHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{e^{\frac{1}{h}} - 2}{e^{\frac{1}{h}} + 2}$. Dividing numerator and denominator by $e^{\frac{1}{h}}$:$RHL = \mathop {\lim }\limits_{h 	o 0^+} \frac{1 - 2e^{-\frac{1}{h}}}{1 + 2e^{-\frac{1}{h}}} = \frac{1 - 0}{1 + 0} = 1$.Since $LHL 
eq RHL$,the limit does not exist.

$\mathop {\lim }\limits_{x \to 1} f(x)$ is equal to,where $f(x) = \begin{cases} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2} & x \neq 1 \\ 1 & x = 1 \end{cases}$

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}$,where $a+b+c \neq 0$.

$\mathop {\lim }\limits_{x \to {2^ + }} \frac{{1 - \cos \{ {x^2} + 2x\} }}{{\ln {{(x - 1)}^{(x - 2)}}}}$ is equal to (where $\{.\}$ denotes fractional part function).

$\mathop {\lim }\limits_{x \to 0} \frac{{x({2^x} - 1)}}{{1 - \cos x}} = $

The value of the limit $\lim_{x \rightarrow 1} \frac{\sin(e^{x-1}-1)}{\log x}$ is

Assertion $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
Reason $(R)$: As the value of $x$ decreases,the value of $\frac{1}{x}$ increases.

Vedclass Test Series

Exam Paper Generator

Online Exam Module