Find the limit of $\lim_{x\to 1}\frac{\sqrt{1-\cos{2(x-1)}}}{x-1}$.

Question

Find the limit: $\lim_{x\to 1}\frac{\sqrt{1-\cos{2(x-1)}}}{x-1}$.

My solution: Let $t=x-1$ then as $x\to 1$, $t\to 0$.

Therefore $\lim_{x\to 1}\frac{\sqrt{1-\cos{2(x-1)}}}{x-1}= \lim_{t\to 0}\frac{\sqrt{1-\cos2t}}{t}=\lim_{t\to 0}\frac{\sqrt{2\sin^2t}}{t}=\sqrt{2}\lim_{t\to 0}\frac{\sin t}{t}=\sqrt{2}$.

But my teacher said that the limit does not exist. So my problem is what's wrong with my solution? Where did I go wrong? Is there any limit?

Any help will be nice. Thank you.

Answer 1

The issue is $\sqrt{\sin^2 t}\ne \sin t$, but is instead $|\sin t|$. That means is always greater or equal to $0$. So on the posiive side the limit is $\sqrt 2$, while on the negative side is $-\sqrt 2$.

Answer 2

Consider $\lim_{t\rightarrow 0} \sqrt{2\sin^2 t}$, $$\lim_{t\rightarrow 0^+} \sqrt{2\sin^2 t}=\sqrt{2}\sin t\\\lim_{t\rightarrow 0^-} \sqrt{2\sin^2 t}=\sqrt{2}\sin \left(-t\right)=-\sqrt{2}\sin t$$ Then, the limit doesn't exist because of upper bound and lower bound are not the same.

Answer 3

Apart from the good answers, a picture is worth a thousand words.