# Solve limit without L'Hôpital's rule

I am trying to solve the following limit without using L'Hôpital's rule. I know the answer is -8, but I can't seem to figure out another way to solve it.

$$\lim_{x\to1} \frac{x^2 -1}{2-\sqrt{x+3}}$$

First let $x=u+1$.

$$L=\lim_{u\to0}\frac{(u+1)^2-1}{2-\sqrt{u+4}}=\lim_{u\to0}\frac{u(u+2)}{2-\sqrt{4+u}}$$

By binomial expansion, we have

$$\sqrt{4+u}=2+\frac14u-\frac1{64}u^2+\mathcal O(u^3)$$

Thus, we have

$$L=\lim_{u\to0}\frac{u(u+2)}{2-\sqrt{4+u}}=\lim_{u\to0}\frac{u(u+2)}{-\frac14u+\frac1{64}u^2+\mathcal O(u^3)}=\lim_{u\to0}\frac{u+2}{-\frac14+\frac1{64}u+\mathcal O(u^2)}=\frac2{-\frac14}=-8$$

• ._. Now what could be the reason now? – Simply Beautiful Art Mar 18 '17 at 15:49
• I think at the time that this was posted, the OP had plenty of hints and answers. Is this a "show off" answer? I doubt the OP knows your $\mathcal O(u^2)$ notation or why you used it or what it means, or how to use it. And no, I did not down vote you answer! – Namaste Mar 18 '17 at 15:51
• Its just a completely different answer, and I think its good to have a mixed basket. – Simply Beautiful Art Mar 18 '17 at 15:52
• I did not see your answer; so, I deleted mine. I totally agree with your statement it is good to have a mixed basket. If I may suggest : when you use Taylor expansions, use one extra term : this gives the asymptotics. If you want to understand my concerns, have a look at matheducators.stackexchange.com/questions/8339/… . Cheers. – Claude Leibovici Mar 18 '17 at 16:01
• I see nothing wrong with using big O notation. (+1) – Mark Viola Mar 18 '17 at 16:39

multiply numerator and denominator by $$2+\sqrt{x+3}$$

• @SimplyBeautifulArt I down voted - momentarily - just to try to fight the snowball effect that comes when people start up voting. I did it for two reasons: 1) this user has nearly 30K rep., he has no business answering this kind of questions without giving greater insight. Leave it for the low rep. users. 2) Pretty he didn't even make sure his idea works – Git Gud Mar 18 '17 at 15:42
• @GitGud. Not providing complete solutions is celebrated, expecially when the asker hasn't any effort whatsoever. Secondly, to you really have any reason to think this substitution won't work?! I credit Dr.Sonnhard, and lab for not spoon-feeding this "do-my-homework for me" asker. – Namaste Mar 18 '17 at 15:45
• So, @GigGud, I suppose that it's only Zain who answered according to your standard, by using Dr. Sonnhard's Hint, and doing simple calculations to spoon-feed the asker? Some of us like to have the participation of the asker! – Namaste Mar 18 '17 at 15:47
• @amWhy You're arguing against a point I didn't make. I have no problem with hints. I had no reason to think this idea didn't work, but one should always check it actually works instead of rushing to answer. And I down voted Zain temporarily too. – Git Gud Mar 18 '17 at 15:54
• @GitGud Well, I can't argue against how you vote. By the way, still didn't figure out those multivariable limits XD – Simply Beautiful Art Mar 18 '17 at 15:57

HINT:

Set $2-\sqrt{x+3}=u \implies x=(2-u)^2-3,u\to0$

alternatively, $\sqrt{x+3}=v$ $x=v^2-3$ and $v\to?$

We have $$\frac{x^2 - 1}{2 -\sqrt{x+3}} = \frac{(x^2 - 1)(2 + \sqrt{x+3})}{(2 - \sqrt{x+3})(2+\sqrt{x+3})} = \frac{(x-1)(x+1)(2+\sqrt{x+3})}{1-x}$$

So $$\lim_{x\to 1} \frac{x^2 -1 }{2-\sqrt{x+3}} = -\lim_{x\to 1} (x+1)(2+\sqrt{x+3}) = -8$$