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I am trying to evaluate

$$\lim_{x \to 0} \frac{\sin(x^3)}{x}$$

without L'hopital rule. I've tried Squeeze theorem but no luck.

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  • $\begingroup$ Use the Taylor development of $\sin$ near $0$. $\endgroup$ – Nicolas May 18 '15 at 13:12
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    $\begingroup$ Hint: multiply and divide by $ x^2$, what is $\lim_{x\to 0} \frac {\sin x}{x}$? $\endgroup$ – Joaquin Liniado May 18 '15 at 13:13
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Using the inequality $|\sin t| \leq |t|$ we get

$$ \left\vert \frac{\sin (x^3)}{x} \right\vert \leq \left\vert \frac{x^3}{x} \right\vert=x^2.$$

Now you can use the squeeze theorem.

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$\lim_{x \to 0} \frac{\sin(x^3)}{x}=\lim_{x \to 0} x^2\frac{\sin(x^3)}{x^3}\lim_{x \to 0} x^2 \cdot \lim_{x \to 0}\frac{\sin(x^3)}{x^3}=0^2 \cdot 1=0$

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You can use the differential quotient:

$$\begin{align} \lim_{x\to0} \frac{\sin(x^3)}{x} & = \lim_{x\to0} \frac{\sin(x^3)-\sin(0)}{x} \\ & = \partial_x(\sin(x^3))|_{x=0} \\ & = (\cos(x^3) 3x^2)|_{x=0} \\ & = 0 \end{align}$$

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You can also use Taylor expansion -

$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots$$

So you limit becomes $$\lim_{x\to 0}\frac{x^3-\frac{x^9}{3!}+\dots}{x} =0$$

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$\frac{\sin(x^3)}{x} = x² \frac{\sin(x^3)}{x³}$ and $\lim_{X \to 0} \frac{\sin(X)}{X}=1$.

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$$\lim_{x \to 0} \frac{\sin(x^3)}{x}=\lim_{x \to 0} \frac{\sin(x)}{\sqrt[3]x}=\lim_{x \to 0} \frac{\sin(x)}xx^{2/3}=1\cdot0$$

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