limit?

**Showcase_22** · May 16th 2009, 10:54 AM

Find the limit (if it exists) of $\lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}$ .

It's apparent that $\sin \left( \frac{1}{x} \right)$ is discontinuous (I have a proof for that involving the sequence $a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $b_n=\frac{1}{2n \pi -\frac{\pi}{2}}$ ).

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

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

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

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

**HallsofIvy** · May 16th 2009, 11:10 AM

Originally Posted by Showcase_22

It's apparent that $\sin \left( \frac{1}{x} \right)$ is discontinuous (I have a proof for that involving the sequence $a_n=\frac{1}{2n \pi+\frac{\pi}{2}}$ and $b_n=\frac{1}{2n \pi -\frac{\pi}{2}}$ ).

Trying this as $\left(\frac{x}{sin x}\right)\left(\frac{sin(1/x)}{(1/x)}\right)$ 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 $-\lim_{x \rightarrow 0} \frac{x^2}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2}{\sin x}$

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

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

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

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