Proving the limit of trigonometric function using epsilon delta definition.

Question

I was trying to figure out the problem $$\lim_{x\to 0^+} \frac{x}{\sqrt{1-\cos x}}$$ It is a fairly easy one where I wrote $\sqrt{1- \cos x}=|\sqrt{2} \sin (x/2)|$. Since we are talking about the right-handed limit, we can safely write $|\sqrt{2} \sin (x/2)|=\sqrt{2} \sin (x/2)$. Thus the whole expression reduces to $$\lim_{x\to 0^+} \frac{x}{\sqrt{2} \sin (x/2)}$$ Using the standard result $\lim_{x\to 0} \frac{\sin x}{x} =1$, we can prove that $$\lim_{x\to 0^+} \frac{x}{\sqrt{1-\cos x}}= \frac{2}{\sqrt2}$$ Having proven this limit, I tried to prove it using the epsilon-delta(precise) definition. Hence I wrote $$|f(0+h)-L|< \epsilon$$ which in turn must imply $$|0+h-0|<\delta$$ Thus moving ahead we get $$\bigg|\frac{h}{\sqrt{1-\cos h}}-\frac{2}{\sqrt2} \bigg|< \epsilon$$.

But from here I have not been able to deduce the desired result through which we can make $\delta$ a function of $\epsilon$ and hence I have been unsuccessful in proving the limit using the precise definition. Enlighten me!

Answer 1

Now since $\lim_{x\to 0} \frac{x}{\sin(x)} =1$ we may pick $\delta(\epsilon) >0$ such that $|\frac{\frac{x}{2}}{\sin(\frac{x}{2})} -1|< \frac{\sqrt{2}}{2}\epsilon$.

Fix $\epsilon >0$. Let $x>0$ with $|x|<\delta(\epsilon)$ and note (as you did) that:

\begin{align*} \left| \frac{x}{\sqrt{1-\cos(x)}}-\frac{2}{\sqrt{2}}\right| = \left|\frac{2}{\sqrt{2}}\frac{\frac{x}{2}}{\sin(\frac{x}{2})}-\frac{2}{\sqrt{2}} \right| = \frac{2}{\sqrt{2}}\left|\frac{\frac{x}{2}}{\sin(\frac{x}{2})} -1 \right|<\frac{2}{\sqrt{2}} \frac{\sqrt{2}}{2}\epsilon = \epsilon, \end{align*} as required. So in essence you did all the hard work.