Solve limit without L'Hôpital's rule

Question

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.

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

Thanks in advance.

score 4 · Answer 1 · answered Mar 18 '17 at 15:37

4

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

answered Mar 18 '17 at 15:37

Dr. Sonnhard Graubner

95,283

1

@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
2

@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. – amWhy 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! – amWhy 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
@GitGud Given this question, it shouldn't have taken any answerer any more than two minutes to check out/confirm the suggested substitution. You said I was arguing against a point you didn't make. Well you were arguing based on an assumption that the answerer here didn't check his substitution and assumed, instead, that he rushed to answer. The two actions aren't as diametrically opposed as you suggest. One can see the validity of a suggested substitution , and answer quickly. – amWhy Mar 18 '17 at 16:25

Simply Beautiful Art · Accepted Answer · 2017-03-18T16:04:25.403

4

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

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

edited Mar 18 '17 at 16:04

answered Mar 18 '17 at 15:46

Simply Beautiful Art

74,685

._. 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! – amWhy 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 deleted the comment...not appropriate on the main site :-) Sorry for that. – amWhy Mar 18 '17 at 16:01
2

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 http://matheducators.stackexchange.com/questions/8339/teaching-limits-and-asymptotics-at-the-same-time . Cheers. – Claude Leibovici Mar 18 '17 at 16:01
@ClaudeLeibovici Ah, okay then :-) – Simply Beautiful Art Mar 18 '17 at 16:03
@ClaudeLeibovici I imagine you prefer the long division step too? – Simply Beautiful Art Mar 18 '17 at 16:08
Not at all but it is more more pedagogical for starting. May I confess that I (almost) never use(d) it ? – Claude Leibovici Mar 18 '17 at 16:09
@ClaudeLeibovici Okay :-) I don't like doing the long division unless its necessary for the limit (multiple rational functions and all that) – Simply Beautiful Art Mar 18 '17 at 16:10
2

I see nothing wrong with using big O notation. (+1) – Mark Viola Mar 18 '17 at 16:39

score 3 · Answer 3 · answered Mar 18 '17 at 15:38

3

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 $v\to?$

answered Mar 18 '17 at 15:38

lab bhattacharjee

274,582

score 3 · Answer 4 · answered Mar 18 '17 at 15:40

3

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

answered Mar 18 '17 at 15:40

Zain Patel

16,802

Solve limit without L'Hôpital's rule

4 Answers4