# How do I solve $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ indeterminate limit without the L'hospital rule?

I've been trying to solve this limit without L'Hospital's rule because I don't know how to use derivates yet. So I tried rationalizing the denominator and numerator but it didn't work.

$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$

What is wrong with $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}-1+1}{\sqrt{1+x}-1} = \lim_{x \to 0} \frac{\sqrt{1+x}-1}{\sqrt{1+x}-1} + \lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ = 1 + DIV?

• What's DIV? Please elaborate. Sep 24, 2015 at 15:54
• @Gummybears: I suppose DIV stands for the claim that this term is divergent which is certainly not true. See my answer below. Sep 24, 2015 at 16:01
• @Tintarn I know the answer :P Sep 24, 2015 at 16:03
• Hint $\sqrt {1+y}=1+(y/2)(1+ k(y))$ where $\lim_{y \to 0} k(y)=0,$ Sep 24, 2015 at 16:29

HINT 1:

\begin{align} \sqrt{1+x}-\sqrt{1-x^2}&=\sqrt{1+x}\left(1-\sqrt{1-x}\right)\\\\ &=\sqrt{1+x}\,\,\left(\frac{x}{1+\sqrt{1-x}}\right) \end{align}

HINT 2:

\begin{align} \frac{1}{\sqrt{1+x}-1}&=\frac{\sqrt{1+x}+1}{x} \end{align}

• The first one beaten me. Sep 24, 2015 at 16:30
• @deadbeef Don't worry. You'll be fine. ;-)) ... And thank you for the "best vote!!!" Sep 24, 2015 at 16:31
• Once again, clever answers to simple problems. Lot to learn. +1 Sep 25, 2015 at 3:46
• @Shailesh Wow! Thank you so much for your compliment and vote up!! Each of us has a lot to learn. And the more one learns, the more that becomes apparent. Sep 25, 2015 at 3:49

Notice, use rationalization as follows $$\lim_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$$ $$=\lim_{x\to 0}\frac{1+x-(1-x^2)}{1+x-1}\times \left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$ $$=\lim_{x\to 0}\frac{x^2+x}{x}\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$ $$=\lim_{x\to 0}(x+1)\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$ $$=(0+1)\left( \frac{\sqrt{1+0}+1}{\sqrt{1+0}+\sqrt{1-0}}\right)$$ $$=1\left(\frac{2}{2}\right)=\color{red}{1}$$

Set $x=\cos2y$

$$\lim_{x \to 0}\dfrac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1} =\lim_{y\to\pi/4}\dfrac{\sqrt2\cos y-\sin2y}{\sqrt2\cos y-1}$$

$$=-\sqrt2\lim_{y\to\pi/4}\cos y\cdot\lim_{y\to\pi/4}\dfrac{\sin y-\sin\dfrac\pi4}{\cos y-\sin\dfrac\pi4}$$

Method$\#1:$ $$\lim_{y\to\pi/4}\dfrac{\sin y-\sin\dfrac\pi4}{\cos y-\sin\dfrac\pi4} =\dfrac{\lim_{y\to\pi/4}\dfrac{\sin y-\sin\dfrac\pi4}{y-\dfrac\pi4}}{\lim_{y\to\pi/4}\dfrac{\cos y-\cos\dfrac\pi4}{y-\dfrac\pi4}} =\dfrac{\dfrac{d(\sin y)}{dy}_{(\text{ at } y=\pi/4)}}{\dfrac{d(\cos y)}{dy}_{(\text{ at } y=\pi/4)}}$$

Method$\#2:$ Use Prosthaphaeresis Formulas to get

$$\dfrac{\sin y-\sin\dfrac\pi4}{\cos y-\sin\dfrac\pi4}=\cdots=-\cot\dfrac{y+\pi/4}2$$

• I guess the only possible explanation for why this wonderful answer was downvoted is because the OP apparently asked for what was wrong with his solution, rather than asking for a new approach. Anyway, that's exactly what I would have done, so $(+1)$. Sep 24, 2015 at 16:43
• Wonderful solution. +1. (Though that's not what the OP asked) Sep 25, 2015 at 3:47

Your argumentation is completely right until in the last step you claim that the second term is divergent which is not true. You rather have: $$\frac{1-\sqrt{1-x^2}}{\sqrt{1+x}-1}=\frac{x^2(\sqrt{1+x}+1)}{x(1+\sqrt{1-x^2}}=x\frac{\sqrt{1+x}+1}{1+\sqrt{1-x^2}}$$ Now, the second factor clearly goes to 1 as $x$ approaches 0 and hence the whole term converges to 0 which makes your original term converge to $1+0=1$.

Note that your idea of rationalizing numerator and denominator was indeed a good approach...

Without too many rationalizations: the term $\sqrt{1+x}$ appears several times. So, set $\sqrt{1+x}=1+t$, that makes $x=2t+t^2$ and $1-x=1-2t-t^2$; then your limit becomes $$\lim_{t\to0}\frac{1+t-(1+t)\sqrt{1-2t-t^2}}{t}= \lim_{t\to0}(1+t)\cdot\lim_{t\to0}\frac{1-\sqrt{1-2t-t^2}}{t}$$ provided the limit of the second factor exists; but $$\lim_{t\to0}\frac{1-\sqrt{1-2t-t^2}}{t}= \lim_{t\to0}\frac{1-(1-2t-t^2)}{t(1+\sqrt{1-2t-t^2})}$$ with the usual rationalization. Can you go on?

• Shouldn't this be the limit as $t \rightarrow 1$? Sep 24, 2015 at 20:28
• @AlfredYerger $t=\sqrt{1+x}-1$; if $x\to0$, then … Sep 24, 2015 at 20:29
• oh! I misread your substitution. Very sorry. Sep 24, 2015 at 21:06
• @AlfredYerger Never mind! Sep 24, 2015 at 21:14