# Evaluate $\lim_{x \to 0} \frac{\sin(x³)}{x}$ without L'hopital rule

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.

• Use the Taylor development of $\sin$ near $0$. – Nicolas May 18 '15 at 13:12
• Hint: multiply and divide by $x^2$, what is $\lim_{x\to 0} \frac {\sin x}{x}$? – Joaquin Liniado May 18 '15 at 13:13

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.

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

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}

You can also use Taylor expansion -

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

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

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

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