If f(x) = frac{e^{1/x} - 1}{e^{1/x} + 1},then | vedclass

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

Question

(D) To find $\lim_{x \rightarrow 0^+} f(x)$,let $x \rightarrow 0^+$,then $\frac{1}{x} \rightarrow \infty$. Thus,$\lim_{x \rightarrow 0^+} \frac{e^{1/x

Vedclass · Accepted Answer

$\lim_{x \rightarrow \infty} f(x) = 0$

Vedclass · Answer

(D) To find $\lim_{x \rightarrow 0^+} f(x)$,let $x \rightarrow 0^+$,then $\frac{1}{x} \rightarrow \infty$. Thus,$\lim_{x \rightarrow 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \lim_{x \rightarrow 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}} = \frac{1 - 0}{1 + 0} = 1$. To find $\lim_{x \rightarrow 0^-} f(x)$,let $x \rightarrow 0^-$,then $\frac{1}{x} \rightarrow -\infty$. Thus,$\lim_{x \rightarrow 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$. Since the left-hand limit and right-hand limit are not equal,$\lim_{x \rightarrow 0} f(x)$ does not exist. For $\lim_{x \rightarrow \infty} f(x)$,$\frac{1}{x} \rightarrow 0$,so $\lim_{x \rightarrow \infty} f(x) = \frac{e^0 - 1}{e^0 + 1} = \frac{1 - 1}{1 + 1} = 0$. Therefore,option $D$ is correct.

If $f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}$,then

Similar Questions

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}} = $

If $S_1 = \sum_{r=1}^{n} r$,$S_2 = \sum_{r=1}^{n} r^2$,and $S_3 = \sum_{r=1}^{n} r^3$,then the value of $\lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$ is

Let $m$ and $n$ be two positive integers greater than $1$. If $\lim_{\alpha \rightarrow 0} \left( \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} \right) = -\left( \frac{e}{2} \right)$,then the value of $\frac{m}{n}$ is

$\lim _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}} = $

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - |x|}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module