How to evaluate $\lim\limits_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$?

Question

I know it can be solved easily with L'H$\mathrm{\hat o}$pital's rule, but I am not allowed to use this rule. What I can use is the definition of function limit, rules of limit algebra (sum, product and quotient), and the composite rule. I am only looking for a hint rather than a full solution. Thank you.

Answer 1

Your question is $$\lim_{x\to0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$$. Now rationalize both the numerator and the denominator. Therefore you will get $$\lim_{x\to0} \frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})}$$. Therefore the above limit will be $$\lim_{x\to 0}\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$$. Therefore, the value of the above limit will be $1$, because $$\lim_{x\to 0} (\sqrt{1+x}+\sqrt{1-x})=2$$.

Answer 2

You can let $x = \cos 2\theta$ then $1 + x = 2\cos^2 \theta$ and $1 - x = 2\sin^2 \theta$.

Now, as $x\to 0$, in terms of $\theta$, we see that $\theta \to \frac{\pi}{4}$. So we need to calculate, $$ \lim_{\theta \to \frac{\pi}{4} } \frac{\sqrt{2\cos^2\theta} - \sqrt{2\sin^2 \theta}}{\cos 2\theta} $$

Generally, $\sqrt{a^2} = |a|$, but in this limit problem we can be sloppy and replace $\sqrt{a^2} = a$ (why?) therefore, we need to calculate the following trigonometric limit, $$ = \lim_{\theta \to \frac{\pi}{4}} \frac{\sqrt{2}(\cos \theta - \sin \theta)}{\cos 2\theta }$$ Now use the double angle-sum identity, $\cos 2\theta = \cos^2\theta - \sin^2 \theta$. Therefore, the limit becomes, $$ \lim_{\theta \to \frac{\pi}{4}} \frac{\sqrt{2}(\cos \theta - \sin \theta)}{\cos^2 \theta - \sin^2\theta } = \lim_{\theta \to \frac{\pi}{4}} \frac{\sqrt{2}(\cos \theta - \sin \theta)}{(\cos \theta + \sin \theta)(\cos \theta - \sin \theta) } = \lim_{\theta \to \frac{\pi}{4}} \frac{\sqrt{2}}{\cos \theta + \sin \theta}$$ Now we can evaluate at $\theta = \frac{\pi}{4}$, thus, we get the answer of, $$ \frac{ \sqrt{2} } { \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = 1$$