# Calculus: Finding limit

$$\lim_{x\to0}\frac{\displaystyle\\x^3}{\sin^3x}.$$

Do I use L'Hopital's rule for this? But I can't seem to find the answer.

• No use the limit $\frac{\sin x }{x}$ and continuity – Alex Oct 23 '16 at 15:42
• "No use...." or "No, use..." – DonAntonio Oct 23 '16 at 15:48
• – Workaholic Nov 1 '16 at 21:32

## 4 Answers

Rather than using de L'Hôpital's rule straight away, notice that you are taking the limit $$\lim_{x \to 0} \left(\frac{x}{\sin x}\right)^3 = \left(\lim_{x \to 0} \frac{x}{\sin x}\right)^3 = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)^{-3}$$ This new limit inside the brackets is often given as a "standard limit", or you could apply de l'Hôpital's rule to it - which should make this a lot easier.

Hint: Since $x \mapsto x^3$ is continuous and $x \mapsto 1/x$ is too, we have $$\lim_{x \to 0 }\frac{x^3}{\sin^3x} = \left( \frac{1}{\color{red}{\lim_{x \to 0}\frac{\sin x}{x}}}\right)^3.$$That limit in red you absolutely must know.

Yes , you can use LHopital:

$$\overset{\text{L'Hopital}}{=}\lim\limits_{x\to 0}\frac{3x^2}{3\sin^2 x \cos x}=\lim\limits_{x\to 0}\frac{1}{ \cos x}\lim\limits_{x\to 0}\frac{x^2}{\sin^2 x }\overset{\text{L'Hopital}}{=}1\cdot\lim\limits_{x\to 0}\frac{2x}{2\sin x \cos x}=1\times 1\lim \limits_{x\to 0}\frac{x}{\sin x}\overset{\text{L'Hopital}}{=}\lim\limits_{x\to 0}\frac{1}{\cos x}=1$$

Applying l'Hôpital's rule:

$$\text{L}=\lim_{x\to0}\frac{x^3}{\sin^3(x)}=\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(x^3\right)}{ \frac{\text{d}}{\text{d}x}\left(\sin^3(x)\right)}=\lim_{x\to0}\frac{x^2}{\cos(x)\sin^2(x)}$$

By the product rule:

$$\text{L}=\lim_{x\to0}\frac{1}{\cos(x)}\cdot\lim_{x\to0}\frac{x^2}{\sin^2(x)}$$

Applying l'Hôpital's rule again:

$$\text{L}=\lim_{x\to0}\frac{1}{\cos(x)}\cdot\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(x^2\right)}{ \frac{\text{d}}{\text{d}x}\left(\sin^2(x)\right)}$$ $$=\lim_{x\to0}\frac{1}{\cos(x)}\cdot\lim_{x\to0}\frac{x}{\cos(x)\sin(x)}=\lim_{x\to0}\frac{1}{\cos^2(x)}\cdot\lim_{x\to0}\frac{x}{\sin(x)}$$

Applying l'Hôpital's rule again:

$$\text{L}=\lim_{x\to0}\frac{1}{\cos^2(x)}\cdot\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(x\right)}{ \frac{\text{d}}{\text{d}x}\left(\sin(x)\right)}=\lim_{x\to0}\frac{1}{\cos^2(x)}\cdot\lim_{x\to0}\frac{1}{\cos(x)}=\lim_{x\to0}\frac{1}{\cos^3(x)}=1$$