# Math Help - Limit Help

1. ## Limit Help

$\lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\sin(5x))))) }{x}
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2. Originally Posted by looseenz2
$\lim_{x\to{0}}\frac{\sin(2\sin(3\sin(4\sin(5x))))) }{x}$ $
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You can either apply L'Hospitals rule or else write

$\frac{\sin(2\sin(3\sin(4\sin 5x)))}{x}=$ $\frac{\sin(2\sin(3\sin(4\sin 5x)))}{2\sin(3\sin(4\sin 5x))}$ $\cdot\frac{2\sin(3\sin(4\sin 5x))}{3\sin(4\sin 5x)}\cdot \frac{3\sin(4\sin 5x)}{4\sin 5x}\cdot \frac{4\sin 5x}{x}$

And now you may want to use that since $\lim_{x\to 0}\frac{\sin x}{x}=1$ , then for any constant $k\,,\,\,\lim_{x\to 0}\frac{\sin kx}{x}=k$ (why?) , and if $f(x)$ is a (differentiable at zero) function s.t. $f(x)\xrightarrow[x\to 0]{}0$ , then $\frac{\sin f(x)}{f(x)}\xrightarrow [x\to 0]{}1$

Tonio