# Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$.

We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + 2x \cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)$$ The derivative of the denominator is $$\cos(x)$$. So, $$\lim\limits_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)} = \lim\limits_{x\rightarrow 0}\displaystyle\frac{\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)}{\cos(x)}$$

Is that right so far?

Thanks for the help in advance. Best Regards, Ahmed Hossam

• The numerator is essentially $\pm x^2$ and the numerator $x$. – Yves Daoust Dec 16 '18 at 20:57
• For all practical purposes this is a duplicate of 1, 2 and 3 proving once again that calculus answering machines think that the site rules don't apply to them. – Jyrki Lahtonen Dec 17 '18 at 7:15
• Well, $\lim\limits_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$ ist not the same as $\lim\limits_{x\to0}{\frac{x^2\cdot\cos\frac{1}{x}}{\sin x}}$ – Ahmed Hossam Dec 20 '18 at 10:04

As we know that $$\lim_{x\to 0}{x\over \sin x}=1$$therefore $$\lim_{x\to 0}{x^2\cos {1\over x}\over \sin x}=\lim_{x\to 0}x\cos{1\over x}=\lim_{u\to \infty}{\cos u\over u}=0$$

We don’t need l’Hopital, indeed by standard limits

$$\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\frac{x \cos\left(\frac{1}{x}\right)}{\frac{\sin(x)}x}\to \frac01=0$$

indeed by squeeze theorem

$$\left|x \cos\left(\frac{1}{x}\right)\right|\le |x|\to0$$

• Thanks for your answer. – Ahmed Hossam Dec 20 '18 at 10:05

Hint: Without using L’Hôpital’s Rule, note that

$$\frac{x^2\cos\big(\frac{1}{x}\big)}{\sin x} = \frac{x}{\sin x}\cdot x\cos\bigg(\frac{1}{x}\bigg)$$

and recall $$\lim_\limits{x \to 0}\frac{\sin x}{x} = 1$$.

• order of zero is higher in Numerator then in denominator hemce 0 – maveric Dec 16 '18 at 18:11
• Thanks for your answer. – Ahmed Hossam Dec 20 '18 at 10:06

With the Taylor power series, $$\sin x= x+o(x)$$ $$\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}=\lim_{x\to0}{x \cos\left(\frac{1}{x}\right)}=0$$ Because $$x\to0$$ and $$\cos(1/x)$$ is bounded from $$-1$$ to $$1$$ as $$x\to0$$

• Thanks for your answer. – Ahmed Hossam Dec 20 '18 at 10:06