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The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$
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Verified Answer
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is equal to 0
$\lim _{x \rightarrow 0} x \sin e^{(1 / x)}$
$\begin{aligned} L H L &=f(0-0)=\lim _{h \rightarrow 0}(-h) \sin e^{(-1 / h)} \\ &=-0 \times \sin \left(e^{-\infty}\right)=-0 \times \sin (0)=0 \end{aligned}$
$=0 \times(\mathrm{a}$ finite number between $-1 \text{ to}+1)$
$$
=0 \quad(\because-1 \leq \sin x \leq 1)
$$
$\because \mathrm{LHL}=\mathrm{RHL}$
$\therefore \lim _{x \rightarrow 0} x \sin \left(e^{1 / x}\right)$ exist and equal to 0
$\begin{aligned} L H L &=f(0-0)=\lim _{h \rightarrow 0}(-h) \sin e^{(-1 / h)} \\ &=-0 \times \sin \left(e^{-\infty}\right)=-0 \times \sin (0)=0 \end{aligned}$
$=0 \times(\mathrm{a}$ finite number between $-1 \text{ to}+1)$
$$
=0 \quad(\because-1 \leq \sin x \leq 1)
$$
$\because \mathrm{LHL}=\mathrm{RHL}$
$\therefore \lim _{x \rightarrow 0} x \sin \left(e^{1 / x}\right)$ exist and equal to 0
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