3
$\begingroup$

The value of $\lim_{x\to \infty} (x+2) \tan^{-1} (x+2) - x\tan^{-1} x $ is $\dots$

a) $\frac{\pi}{2} $ $\qquad \qquad \qquad$ b) Doesn't exist $\qquad \qquad \qquad$ c) $\frac{\pi}{4}$ $\qquad \qquad$ d)None of the above.

Now, this is an objective question and thus, I expect that there must be an easier way to do it either by analyzing the options or something else. I'm not sure about this though!

What I've done yet is trying to apply : $\tan^{-1}a - \tan^{-1}b$ formula but it requires certain conditions to hold true which are not specified here. I'm not sure how we will approach this! Any kind of help will be appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ Hint: Split $(x+2) \tan^{-1}(x+2) = x \cdot \tan^{-1} (x+2) + 2 \tan^{-1} (x+2) $, apply $\tan(A-B) = \frac{\tan A - \tan B}{1 - \tan A \tan B} $, for small $ x$, $ x \approx \tan x $ $\endgroup$
    – GohP.iHan
    Commented May 3, 2015 at 12:39
  • 1
    $\begingroup$ I started by writing $\frac{1}{\tan (x)}$ facepalm $\endgroup$ Commented May 3, 2015 at 15:41

6 Answers 6

4
$\begingroup$

Let $a=\tan^{-1}(x+2),b=\tan^{-1}x$.

Then, since one has $\tan a=x+2,\tan b=x$, one has$$\begin{align}\lim_{x\to \infty}a\tan a-b\tan b&=\lim_{x\to\infty}a(\tan b+2)-b\tan b\\&=\lim_{x\to\infty}2a+(a-b)\tan b\\&=\lim_{x\to \infty}2\tan^{-1}(x+2)+\frac{\tan^{-1}(x+2)-\tan^{-1}x}{\frac 1x}\\&=2\cdot\frac{\pi}{2}+\lim_{x\to\infty}\frac{\frac{1}{1+(x+2)^2}-\frac{1}{1+x^2}}{-\frac{1}{x^2}}\\&=\pi+\lim_{x\to\infty}\left(-\frac{x^2}{1+(x+2)^2}+\frac{x^2}{1+x^2}\right)\\&=\pi-1+1\\&=\pi\end{align}$$

$\endgroup$
3
  • $\begingroup$ I've a question! What did you first think about the question when you saw it? I mean, how did you realize that this method will work. Specially when there is no time to think about which method will work or not, in competitive exams having objective questions, each to be done in 2 - 3 minutes (atmost). $\endgroup$ Commented May 3, 2015 at 12:43
  • $\begingroup$ @KushashwaRaviShrimali: I thought setting $a=\tan^{-1}(x+2),b=\tan^{-1}x$ would be helpful and getting $\tan a=\tan b+2$ helps. $\endgroup$
    – mathlove
    Commented May 3, 2015 at 12:46
  • 1
    $\begingroup$ That is something I've always been looking for, which is, the first thought while solving a question. I usually see the options to see what the asker is demanding for and then I proceed. Thanks a lot for your kind help, Mathlove. I've a few questions left, hope to get help from you there too! $\endgroup$ Commented May 3, 2015 at 12:49
2
$\begingroup$

you can use the mean value theorem on $$f(x) = x\tan^{-1} x, \quad f'(x)= \tan^{-1} x + \frac{x}{1+x^2} \to \pi/2 \text{ as } x \to \infty.$$

by mvt, $$(x+2)f(x+2) - xf(x) = 2f'(x+\epsilon) \to \pi \text{ as } x \to \infty.$$

$\endgroup$
3
  • $\begingroup$ Well to be honest, I've never studied mvt yet. So, it will take a bit time to get comfortable with it. But, still, I hope with such a nice explanation, I will be able to make my way through! Thanks a lot Mr. Abel for helping me out. $\endgroup$ Commented May 3, 2015 at 12:42
  • $\begingroup$ @KushashwaRaviShrimali, you are welcome. mvt is, i believe, one of the corner stones of calculus. take your time. $\endgroup$
    – abel
    Commented May 3, 2015 at 12:44
  • $\begingroup$ Oh, that indeed inspires me to go with it! :) Thanks again. $\endgroup$ Commented May 3, 2015 at 12:45
2
$\begingroup$

Since $$ \tan^{-1}(x)=\frac\pi2-\tan^{-1}\left(\frac1x\right)=\frac\pi2-\frac1x+O\left(\frac1{x^3}\right) $$ we get $$ x\tan^{-1}(x)=\frac\pi2x-1+O\left(\frac1{x^2}\right) $$ Thus, $$ (x+2)\tan^{-1}(x+2)-x\tan^{-1}(x)=\frac\pi2\cdot2+O\left(\frac1{x^2}\right) $$ so that $$ \lim_{x\to\infty}\left[(x+2)\tan^{-1}(x+2)-x\tan^{-1}(x)\right]=\pi $$

$\endgroup$
1
  • $\begingroup$ Awesome! I was just thinking that Taylor Series is one of the technique that comes into play while evaluating the limits. Why not here? :D And it finally arrives! Thanks a lot for sparing time to help me out. $\endgroup$ Commented May 3, 2015 at 12:49
1
$\begingroup$

Here again ! When $y$ is large, an asymtotic expansion is $$\tan^{-1}(y)=\frac{\pi }{2}-\frac{1}{y}+\frac{1}{3 y^3}+O\left(\left(\frac{1}{y}\right)^4\right)$$ $$A=(x+2) \tan^{-1} (x+2) - x\tan^{-1}(x)$$ $$A\approx(x+2)\Big(\frac{\pi }{2}-\frac{1}{x+2}+\frac{1}{3 (x+2)^3}\Big)-x\Big(\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}\Big)=\pi -\frac{4 (x+1)}{3 x^2 (x+2)^2}$$ where you see the limit and how it is approached.

Again, plot the function and the approximation for $0 \leq x\leq 10$ and tell me if this is funny or not.

$\endgroup$
5
  • $\begingroup$ @KushashwaRaviShrimali. You please stop kiding me and forget the Sir. You make me older that I am !! $\endgroup$ Commented May 3, 2015 at 12:56
  • $\begingroup$ That graph almost looked like a heart pulse, with infinite top and bottom. ^^ $\endgroup$
    – Someone
    Commented May 3, 2015 at 13:13
  • $\begingroup$ Oh, I see the edited portion now. That's cool! $\endgroup$ Commented May 3, 2015 at 13:16
  • $\begingroup$ @Mann. I don't see what you mean. Are not the curves very close to eachother ? $\endgroup$ Commented May 3, 2015 at 13:18
  • $\begingroup$ desmos.com/calculator/5c8chh9neh , take a look^^ It's $+\infty$ and $-\infty$ at points and then it's just constant $\pi$! $\endgroup$
    – Someone
    Commented May 3, 2015 at 13:29
1
$\begingroup$

Set $1/x=h$ to get $$\lim_{h\to0^+}\dfrac{(1+2h)\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h}h$$

$$=\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h}h+2\lim_{h\to0^+}\tan^{-1}\dfrac{1+2h}h$$

Now, $$\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h=\tan^{-1}\left[\dfrac{\dfrac{1+2h}h-\dfrac1h}{1+\dfrac{1+2h}h\cdot\dfrac1h}\right]=\tan^{-1}\dfrac{2h^3}{(h+1)^2}$$

$$\implies\lim_{h\to0^+}\dfrac{\tan^{-1}\dfrac{1+2h}h-\tan^{-1}\dfrac1h}h=\lim_{h\to0^+}O(h^2)=0$$

Finally, $$\lim_{h\to0^+}\tan^{-1}\dfrac{1+2h}h=\tan^{-1}(+\infty)=+\dfrac\pi2$$

$\endgroup$
2
  • $\begingroup$ Thanks a lot for your help.. I was wondering, whats basically the difference between $\infty ^{-} $ and $-\infty$ ? $\endgroup$ Commented May 3, 2015 at 13:17
  • $\begingroup$ From that I meant, what difference is there in : $x \to \infty^{-} $ and $ x \to - \infty$ ? $\endgroup$ Commented May 3, 2015 at 13:17
1
$\begingroup$

I saw @mathlove's solution a different way, in case you find it hard to come up with convenient substitutions:

$(x+2)\arctan (x+2) - x \arctan(x)$ = $x(\arctan (x+2) - \arctan(x) )+2\arctan(x)$

Look at the first tirm: $x$ is going to $\infty$, but $\arctan (x+2) - \arctan(x)$ is going to $0$. And if you know the graph of $\arctan(x)$, you know the difference is going to $0$ "pretty quickly"--which is just intuition to help us know that the product is probably going to 0.

We can prove it goes to 0 with L'Hopital:

$x(\arctan (x+2) - \arctan(x))=\frac{\arctan (x+2) - \arctan(x)}{\frac{1}{x}}$,

So

$\lim\limits_{x\rightarrow \infty} x(\arctan (x+2) - \arctan(x))=\lim\limits_{x\rightarrow\infty} \frac{\frac{1}{1+(x+2)^2}-\frac{1}{1+x^2}}{-\frac{1}{x^2}}=0$

because the $x^4$ terms cancel out of the simplified numerator.

That leaves us with $2\arctan(x)$, but it's easy to see this goes to $\pi$.

Basically the key was I saw $x$ multiplying two $\arctan$ terms, so I had the instinct to factor them. Add in some good productive staring, and voila.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .