lim _{x rightarrow 0}left(frac{sinh 2 x}{2 x} | vedclass

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

Question

(C) Let $L = \lim _{x \rightarrow 0}\left(\frac{\sinh 2 x}{2 x}\right)^{\frac{1}{x^2}}$.Using the Taylor series expansion for $\sinh(u) = u + \frac{u^

Vedclass · Accepted Answer

$e^{2/3}$

Vedclass · Answer

(C) Let $L = \lim _{x \rightarrow 0}\left(\frac{\sinh 2 x}{2 x}\right)^{\frac{1}{x^2}}$.Using the Taylor series expansion for $\sinh(u) = u + \frac{u^3}{6} + O(u^5)$,we have $\sinh(2x) = 2x + \frac{(2x)^3}{6} + O(x^5) = 2x + \frac{4x^3}{3} + O(x^5)$.Thus,$\frac{\sinh 2x}{2x} = 1 + \frac{2x^2}{3} + O(x^4)$.Now,$L = \lim _{x \rightarrow 0} \left(1 + \frac{2x^2}{3} + O(x^4)\right)^{\frac{1}{x^2}}$.Using the standard limit $\lim _{u \rightarrow 0} (1+u)^{1/u} = e$,let $u = \frac{2x^2}{3}$. As $x \rightarrow 0$,$u \rightarrow 0$.Then $L = \lim _{x \rightarrow 0} \left[ \left(1 + \frac{2x^2}{3}\right)^{\frac{3}{2x^2}} \right]^{\frac{2}{3}} = e^{2/3}$.

$\lim _{x \rightarrow 0}\left(\frac{\sinh 2 x}{2 x}\right)^{\frac{1}{x^2}} = $

Similar Questions

$\mathop {\lim }\limits_{x \to 1} \frac{x - 1}{2x^2 - 7x + 5} = $

$\mathop {\lim }\limits_{x \to 3} \left\{ {\frac{{x - 3}}{{\sqrt {x - 2} - \sqrt {4 - x} }}} \right\} = $

The value of $\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$ is

$\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1} = $

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of $\mathop {\lim }\limits_{x \to 1} (1 - x + [x - 1] + [1 - x])$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module