# Proving the limit of trigonometric function using epsilon delta definition.

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!

• Well every limit should not be proved using the definition of limit. In order to use the $\epsilon - \delta$ approach you will need to use the standard limit \lim_{x\to 0}\dfrac{\sin x} {x}=1$here. You can't avoid that. – Paramanand Singh Apr 24 '17 at 13:36 • You might want to take a look at this related question: math.stackexchange.com/questions/285313/… – Guangliang Apr 24 '17 at 13:39 • Moreover your idea that an inequality of type$|f(x) - L|<\epsilon$should imply an inequality of type$|x-a|<\delta$is wrong. The limit definition does not work like that. The central idea is that for given$\epsilon >0$it should be possible to find a$\delta>0$such that the implication$0<|x-a|<\delta\Rightarrow |f(x) - L|<\epsilon$can be ensured. Note that the implication can not in general be done via chain of algebraic manipulation to convert one inequality into another. – Paramanand Singh Apr 24 '17 at 13:41 ## 1 Answer 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.