Evaluating $\lim_{x\to\infty}\left(\frac{\sinh x}{x}\right)^{{1}/{x^2}}$

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$$\lim_{x\to\infty}\left(\frac{\sinh x}{x}\right)^{\dfrac{1}{x^2}}$$

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Hint: Use Squeezing. Note that $1\lt \frac{\sinh x}{x}\lt e^x$ when $x$ is large.