$$\lim _{ x\rightarrow 1 }{ (x-1) } \sin { \frac { \pi }{ x-1 } } $$

Steps I took:

$$-1\le \sin { \frac { \pi }{ x-1 } } \le 1$$

$$-\left| x-1 \right| \le \sin { \frac { \pi }{ x-1 } } \le \left| x-1 \right| $$

$$\lim _{ x\rightarrow 1 }{ -\left| x-1 \right| } =0$$

$$\lim _{ x\rightarrow 1 }{ \left| x-1 \right| } =0$$

$$\lim _{ x\rightarrow 1 }{ (x-1)\sin { \frac { \pi }{ x-1 } } } =0$$

My final answer is correct according to the answer key of the book I am using, but quite frankly, I have no idea why I took any of the steps that I did, if any of them are even correct, and why this has to be done the way it is. I like to understand the underlying reason for the steps I take when solving a problem. Khan Academy usually does a great job of breaking it all down for me but unfortunately this concept isn't covered on there. I am hoping to get an explanation for this so that I can solve other problems like this and know what to do and why I have to do it that way.

  • $\begingroup$ This is correct, the squeeze theorem is ideal for things like sine that oscillate, that's the motivation, since the other limits, i.e. $|x-1|$ as $x\to 1$ are easy. $\endgroup$ – Adam Hughes Jan 20 '15 at 2:35
  • $\begingroup$ Small typo, second line, need to multiply middle term by $|x-1|$. $\endgroup$ – André Nicolas Jan 20 '15 at 2:36
  • $\begingroup$ Would you please be able to explain why? $\endgroup$ – Cherry_Developer Jan 20 '15 at 2:38
  • 1
    $\begingroup$ As $x$ approaches $1$, $|x-1|\sin\left(\frac{\pi}{x-1}\right)$ is free to wiggle between the walls $-|x-1|$ and $|x-1|$. However, these walls are both approaching $0$, so the poor thing gets crushed in between, and is forced to approach $0$. $\endgroup$ – André Nicolas Jan 20 '15 at 2:43
  • 1
    $\begingroup$ The point of the argument is that the sin term never gets very big, positive or negative. So when it gets multiplied by $x-1$, which goes to $0$, the product goes to $0$. And yes, the inequalities in Line 1 get "multiplied" by $x-1$. Equivalently we could have said in the second line that $-|x-1|\le (x-1)\sin(\pi/(x-1))\le |x-1|$. $\endgroup$ – André Nicolas Jan 20 '15 at 2:56

Taking absolute values makes things easier: $0 \leq | (x-1)\sin (\pi/(x-1)) \leq |x-1|$. As $x \to 1$ both extremities go to zero and therefore so does the middle (by the squeeze theorem).


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