How to find $\lim\limits_{x \to 0} \frac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$?

Question

How to find $\;\;\lim\limits_{x \to 0} \dfrac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$ ?

Hint: Rationalise the denominator and use limx->0 sinx/x=1 and tanx/x=1 — Mr. Math, Nov 29 '14 at 15:11

score 4 · Accepted Answer · answered Nov 29 '14 at 15:12

4

Hint: Multiply the numerator and denominator by $\sqrt{1 + \tan x} + \sqrt {1-\tan x}$.

Your denominator will become $1 + \tan x - (1-\tan x) = 2\tan x$.

answered Nov 29 '14 at 15:12

amWhy

209,954

4

This is a good answer: good hints, no spoon-feeding. +1 – Timbuc Nov 29 '14 at 15:17
1

@Timbuc Well, someone out there has been taking delight in downvoting my answers lately. Thank you for your vote of confidence. – amWhy Nov 29 '14 at 15:20
1

I haven't seen him in a while, but I'm sure my teacher donantonio sends you his best regards. He also used to tell me, when he introduced me to this site and taught me how to write and etc., that sometimes there are trolling downvoters. It hasn't happened to me so far, but I guess it'll happen sometime in the future. Best Regards. – Timbuc Nov 29 '14 at 15:23

score 2 · Answer 2 · answered Nov 29 '14 at 15:13

2

We have $$f(x) = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} = \dfrac{\sin(2x)}{\sqrt{1+\tan(x)} - \sqrt{1-\tan(x)}} \dfrac{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}{\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}}$$ Hence, $$f(x) = \dfrac{\sin(2x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)}{1+\tan(x) - 1 + \tan(x)} = \dfrac{2\sin(x)\cos(x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)}{2\tan(x)}$$ This gives us $$f(x) = \cos^2(x) \left(\sqrt{1+\tan(x)} + \sqrt{1-\tan(x)}\right)$$ Now take the limit as $x \to 0$ to get $f(x) \to 2$.

answered Nov 29 '14 at 15:13

Adhvaitha

5,441

4

Avoid spoon feeding – lab bhattacharjee Nov 29 '14 at 15:16
1

@labbhattacharjee Don't worry, I used the hint :D – omidh Nov 29 '14 at 15:31
@omidh, Its not only about you. Also, here in MSE, the common practice for answering is to leave some room for thought for the OP – lab bhattacharjee Nov 29 '14 at 15:33
@labbhattacharjee I'm agree with you – omidh Nov 29 '14 at 15:38

Claude Leibovici · Answer 3 · 2014-11-29T15:32:23.993

1

Sinc you received already good answers, let me show you what you can do using Taylor series. You know that, for small values of $x$, $\tan(x)\approx x$ and that $\sin(x)\approx x$. So the expression is almost $$\frac{2x}{\sqrt{1+x}-\sqrt{1-x}}$$ Now remember that $\sqrt{1+y}\approx \frac y2$. So the expression is $$\frac{2x}{(\frac x2)-(-\frac x2)}$$

edited Nov 29 '14 at 15:32

answered Nov 29 '14 at 15:22

Claude Leibovici

260,315

Good for guessing the answer but to make it rigorous you need to include the error terms and throw them away only when not needed. – user21820 Nov 29 '14 at 15:36
Very useful way, Thanks! – omidh Nov 29 '14 at 15:42
@user21820. I totally agree with you but I did not want to invade the space with that. I limited to the strict minimum. May I confess that I m in love with Taylor series ... for more than 55 years ? – Claude Leibovici Nov 29 '14 at 15:43
@omidh. The beauty of Taylor series is that not only you get the limit but you also see how the limit is approached. In the case of your problem, the expression behaves just as $2-\frac{9 x^2}{4}+O\left(x^4\right)$. Plot the two curves on the same graph for $-1/2<x<1/2$; you should be amazed. – Claude Leibovici Nov 29 '14 at 15:48
@ClaudeLeibovici: Did you ever see http://math.stackexchange.com/a/809591/21820? You should be able to get the answer without looking at my solution. Some computer algebra systems cannot. – user21820 Nov 29 '14 at 15:50
@user21820. This is incredibly beautiful ! For sure, for this kind of problems, CAS are just dumb ! Thank you. Cheers :-) – Claude Leibovici Nov 29 '14 at 15:54

How to find $\lim\limits_{x \to 0} \frac{\sin 2x}{\sqrt{1+\tan x} - \sqrt{1-\tan x }}$?

3 Answers3