Evaluate the limit or show that the limit does not exist for n ∈ Z in $ (x − 1)^n \sin (\frac1{x − 1})$

Question

Evaluate the limit or show that the limit does not exist for the function below.$$\lim_{x→1} (x − 1)^n \sin \left(\frac{1}{x − 1}\right)$$

{How to start this question? we haven't been taught L'Hopital's rule yet.

What I have done so far is:

let $ \alpha = \frac{1}{1-x}$. $\;$

Then $ -1\le \sin \alpha \le 1$. $\Rightarrow$ $\; -(x-1)^n\le (x − 1)^n\sin \alpha \le (x-1)^n$. $\;$ Taking limits on both sides by squeeze theorem we have $\lim_{x→1} -(x − 1)^n =\lim_{x→1} (x − 1)^n = 0$

Therefore $$\lim\limits_{x→1} (x − 1)^n \sin \left(\frac{1}{x − 1}\right) =0$$

Is this correct?

Answer 1

Note that this problem is the same as evaluating $$\lim_{x \to 0} x^n\sin(1/x).$$ Suppose that $n > 0$. Let $\epsilon > 0$. Let $\delta = \sqrt[n]{\epsilon}$ if $\epsilon > 1$ and let $\delta = \epsilon$ if $\epsilon \leq 1$. Then $\vert x \vert^n < \epsilon$ whenever $\vert x \vert < \delta$, and noting that $\vert \sin(1/x) \vert \leq 1$, we have $$\vert x^n\sin(1/x) \vert = \vert x \vert^n \vert \sin(1/x) \vert < \epsilon$$ whenever $\vert x \vert < \delta$. Therefore the limit is $0$.

If $n \leq 0$ then the limit doesn't exist. If $n < 0$ you can show the function is unbounded in any neighborhood of $0$ by considering the function at $\pm 2/\pi, \pm 2\pi/3, \pm 2\pi/5, \ldots$. If $n = 0$, note that $\sin(1/x)$ oscillates between $-1$ and $1$ as $x \to 0$, so the limit doesn't exist. Try finishing this up formally.