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

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(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

$e^{2/3}$

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(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}} = $

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