1. ## limit?

Find the limit (if it exists) of $\displaystyle \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}$.
It's apparent that $\displaystyle \sin \left( \frac{1}{x} \right)$ is discontinuous (I have a proof for that involving the sequence $\displaystyle a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $\displaystyle b_n=\frac{1}{2n \pi -\frac{\pi}{2}}$).

Define $\displaystyle -\lim_{x \rightarrow 0} \frac{x^2}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2}{\sin x}$

$\displaystyle \lim_{x \rightarrow 0} \frac{x^2}{\sin x}=\lim_{x \rightarrow} \frac{2x}{\cos x}=0$

Hence $\displaystyle \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}=0$

Is this the correct way of doing this type of question or is there an easier way?

2. Originally Posted by Showcase_22
It's apparent that $\displaystyle \sin \left( \frac{1}{x} \right)$ is discontinuous (I have a proof for that involving the sequence $\displaystyle a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $\displaystyle b_n=\frac{1}{2n \pi -\frac{\pi}{2}}$).
Trying this as $\displaystyle \left(\frac{x}{sin x}\right)\left(\frac{sin(1/x)}{(1/x)}\right)$ should make it much easier! As x goes to 0, 1/x goes to infinity while sin(1/x) is always between -1 and 1.

Define $\displaystyle -\lim_{x \rightarrow 0} \frac{x^2}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2}{\sin x}$

$\displaystyle \lim_{x \rightarrow 0} \frac{x^2}{\sin x}=\lim_{x \rightarrow} \frac{2x}{\cos x}=0$

Hence $\displaystyle \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}=0$

Is this the correct way of doing this type of question or is there an easier way?