1
$\begingroup$

Find the limit: $$\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$$ After treating it with l'Hopital rule, we get: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}-\cos\frac{1}{x}}{\cos x}}$$ Now, the numerator of fraction doesn't have a limit, so I can't use l'Hopital rule again. What to do? I can split it into two fractions, but I'm not sure how would it help: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}}{\cos x}} - \lim_{x\to0}{\frac{\cos\frac{1}{x}}{\cos x}}$$

$\endgroup$
2
$\begingroup$

$$0\leq \left| \frac{x^2 \text{sin}\frac{1}{x}}{\text{sin} x} \right| \leq \left| \frac{x^2}{\sin x} \right|$$

You can now apply L'Hôpital's rule to $\frac{x^2}{\text{sin}x}$ to show the limit of this as $x \rightarrow 0$ is $0$, then use the squeeze rule.

$\endgroup$
  • $\begingroup$ Thanks for the edit! Enlightened me to \sin - I've always used \textrm - sigh... $\endgroup$ – ah11950 Mar 26 '14 at 1:09
2
$\begingroup$

You can only split if both limits exist. In this case, it's fine. However, l'Hospital doesn't help here. The derivative of $\sin \frac{1}{x}$ wildly diverges so it's useless. What you need to realize is, that $x\sin\frac{1}{x}$ is a product of a bounded function and a function with limit $0$, which tells you about the limit.

The part $\frac{x}{\sin x}$ converges to 1, that's known.

$\endgroup$
1
$\begingroup$

Rewrite the function as $$ \frac{\left(\frac{\sin \frac1x}{\frac1x}\right) }{\left(\frac{\sin x}{x}\right)} $$ Now think about the behavior of the simpler function $g(x)=\sin x/ x$ both as $x\rightarrow 0$ and as $x\rightarrow \pm \infty$.

As $x\rightarrow 0$, $g(x) \rightarrow 1$.

As $x\rightarrow \pm\infty$, $g(x)\rightarrow 0$.

These are precisely the behaviors of the denominator and numerator, respectively, as $x\rightarrow 0$. So the limit of the original expression as $x\rightarrow 0$ is $0/1=0$.

$\endgroup$
0
$\begingroup$

$$ \lim_{x\to 0} \frac{x^2\sin\left(\frac{1}{x}\right)}{\sin(x)} = \left[\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)\right]\left[\lim_{x\to 0}\frac{x}{\sin(x)}\right]= \lim_{x\to 0} x\sin\left(\frac{1}{x}\right) $$ Let $z=\frac{1}{x}$. Since $x\to 0$, then $z\to\infty$. So now we have $$ \lim_{z\to \infty} \frac{\sin\left(z\right)}{z} =0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.