Limit of (x^2 - 1)sin(1/(x-1)) as x approaches 1

Can anybody show me how to proceed with this question?

$\text{Evaluate} \ \displaystyle \lim_{x \to 1} \left( \left(x^2 - 1 \right) \sin \left( \frac{1}{x-1} \right) \right)$

So far I've noticed that:

$\displaystyle \lim_{x \to 1} \left(x^2 - 1 \right) = 0$

However, I'm unsure of how to proceed with the limit of the sine function.

We have xsin(1/x) -> 0 as x->0 (because it's bounded by |x| in absolute value). Use x^2 - 1 = (x+1)(x-1).


Posted from TSR Mobile

May I ask how I'm supposed to use the first part? Also I don't understand what's written in your parentheses.

Sure.

For what I put in brackets: we have 0 =< |xsin(1/x)| = |x| |sin(1/x)| =< |x| because |sin(u)| =< 1 for every u. So, since |x| -> 0 as x -> 0, we get xsin(1/x) -> 0 too.

Then, by the above, (x-1)sin(1/(x-1)) -> 0 as x -> 1.

Now: (x^2 - 1)sin(1/(x-1)) = (x+1) (x-1)sin(1/(x-1)). Combining the above with the fact that (x+1) is bounded for x near 1, you should be able to get the answer.

According to what you've written down (all of with which I agree),

$\displaystyle \begin{aligned} \lim_{x \to 1} \ \left[ (x^2 - 1) \sin \left( \frac{1}{x-1} \right) \right] & = \lim_{x \to 1} \ \left[ (x+1) (x-1) \sin \left( \frac{1}{x-1} \right) \right] \\ & = \lim_{x \to 1} \ (x+1) \cdot \lim_{x \to 1} \ \left[ (x-1)\sin \left( \frac{1}{x-1} \right) \right] \\ & = 2 \times 0 = 0 \end{aligned}$

Is this correct?

I made the hand wavey argument that $\forall x: \ -1 \leq \sin(x) \leq 1$

According to what you've written down (all of with which I agree),

$\displaystyle \begin{aligned} \lim_{x \to 1} \ \left[ (x^2 - 1) \sin \left( \frac{1}{x-1} \right) \right] & = \lim_{x \to 1} \ \left[ (x+1) (x-1) \sin \left( \frac{1}{x-1} \right) \right] \\ & = \lim_{x \to 1} \ (x+1) \cdot \lim_{x \to 1} \ \left[ (x-1)\sin \left( \frac{1}{x-1} \right) \right] \\ & = 2 \times 0 = 0 \end{aligned}$