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I was trying to solve this problem, which says to find the limit:

$$\lim_{x\to\infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$$

Since the OP said the answer was one, and I knew that $\displaystyle\lim_{x\to\infty} x^{\large \frac 1x}=1$ , I wanted to show that $$\lim_{x\to\infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)=\lim_{x \to\infty}x\;\;\;\text{or}\;\;\;\lim_{x\to\infty} \large \frac {\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}=1$$ .

Using this assumption I arrived at $$\lim_{x\to\infty}\cot x =\lim_{x\to\infty}x,$$ which is nonsense because the limit on the LHS does not exist. When I later checked what this limit was on Wolframalpha, I found that $\displaystyle\lim_{x\to\infty} \large \frac {\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}=0$ . So how come that since this limit is not $1$ , $\displaystyle\lim_{x\to\infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$ is $1$? Thank you in advance.

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  • $\begingroup$ Actually I think I just figured this out. $\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$ is actually equal to $\lim_{x \to \infty} \frac 1x$ , and $\lim_{x \to \infty} (1/x)^{(1/x)}=1$ $\endgroup$
    – Ovi
    Jul 11, 2013 at 2:33

4 Answers 4

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Look at the function $\arctan(x)$. It tends towards $\pi/2$ as $x \rightarrow \infty$. So the quantity in parentheses in your first equation goes towards zero, but since the exponent is also going to zero, you get 1. Meanwhile, the function $x$ obviously goes to infinity in the limit. This is all basically a result of the fact that when you raise anything to the 0 power, you get 1. Similarly, just because $5^0=7^0$, you cannot conclude that $5=7$.

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  • $\begingroup$ "but since the exponent is also going to zero, you get 1" - ? $\endgroup$
    – anon
    Jul 11, 2013 at 2:40
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There are a lot of issues in what you write: mainly, if you know that $\lim_{x\to\infty} f(x) = \lim_{x\to\infty} g(x)$, you cannot conclude from it that $\lim_{x\to\infty} f(x)/g(x) = 1$.

For example, $\lim_{x\to\infty} \frac{n^2}{n} \neq 1$, and $\lim_{x\to\infty} \frac{n}{n^2} \neq 1$.

So let's go through your question step-by-step:

  • Firstly, because $\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}} = 1$, and $\lim_{x \rightarrow \infty} x^{\large \frac 1x}=1$, you want to show that $\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)=\lim_{x \to \infty}x$. This is already wrong: the left-hand-side limit is $0$, while the right-hand-side limit is $\infty$. (What would be correct is wanting to show that $\displaystyle \lim_{x\to\infty} \frac{\left(\frac{\pi}{2}-\tan^{-1}x\right)^{1/x}}{x^{1/x}} = 1$, but you cannot get rid of the $\frac1x$ in the exponent like that.)

  • Next, even to show that $\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)=\lim_{x \to \infty}x$, you want to show that $\lim_{x \rightarrow \infty} {\large \frac {\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}}=1$: wrong idea again, because $\lim_{x\to\infty} f(x) = \lim_{x\to\infty} g(x)$ may be true without it being true that $\lim_{x\to\infty} f(x)/g(x) = 1$.

  • From this you arrived (again I'm not exactly sure how) at the conclusion that $\lim_{x \to \infty} \cot x =\lim_{x \to \infty} x$, which you yourself realized is nonsense.

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  • $\begingroup$ I know that if two functions have the same limit their ration is not necessarily $1$ , but in a cases where $\lim_{x \to \infty} (f(x))^{1/x}=1$ , you could argue that in the cases where $\lim_{x \to \infty} f(x) \not =1/x$ , then $\lim_{x \to \infty} \frac {f(x)}{x}=1$ , right? $\endgroup$
    – Ovi
    Jul 11, 2013 at 2:49
  • $\begingroup$ And the way I arrived at $\lim_{x \to \infty} \cot x =\lim_{x \to \infty} x$ is by treating my limit as an equation, solving for $\arctan x , and taking the tangent of both sides$ . Is this wrong too? I know that sometimes you can and sometimes you can't treat a limit like an equation. $\endgroup$
    – Ovi
    Jul 11, 2013 at 2:52
  • $\begingroup$ @Ovi: Wrt the first comment: no it's not right that if $\lim_{x\to\infty}(f(x))^{1/x}=1$ and $\lim_{x\to\infty} f(x)\neq1/x$, then $\lim_{x\to\infty}\frac{f(x)}{x}=1$: this particular one is itself an example, with $f(x)=\frac{\pi}{2}-\tan^{-1}x$ and $\lim_{x\to\infty} f(x) = 0$. In fact the limit $\lim_{x\to\infty}(f(x))^{1/x}=1$ holds for $f(x)$ being a lot of things: $x$, $x^2$, $x^3$, $x^n$, $x^{0.5}$, $\log x$, $\log \log x$, $1$, $2$, $\frac1x$... basically, intuitively $f(x)$ growing at any rate slower than exponential in $x$. (For $f(x)=c^x$, we have $\lim_{x\to\infty}(f(x))^{1/x}=c$.) $\endgroup$ Jul 11, 2013 at 3:16
  • $\begingroup$ @Ovi: Wrt to the second, for the precise details you need to look up a textbook, but roughly you can do some things (treating it as an "equation") when the limits involved are not infinite. So you have things like $\lim f(x) = \lim g(x) \implies \lim f(x)/g(x) = 1$ if $\lim f(x) \neq \infty$ (and $\neq 0$). See en.wikipedia.org/wiki/Limit_of_a_function#Properties $\endgroup$ Jul 11, 2013 at 3:18
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Why would you expect that

$$\lim_{x\to\infty}\frac{\frac{\pi}2-\tan^{-1}x}x=1\;?\tag{1}$$

You know that $$\lim_{x\to\infty}\tan^{-1}x=\frac{\pi}2\;,$$

so the numerator of the fraction in $(1)$ approaches $0$ while the denominator blows up. That clearly forces the fraction to approach $0$.

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  • $\begingroup$ I wasn't thinking at all, I was just going on the fact that $\lim_{x \to \infty} x^{1/x}=1$ $\endgroup$
    – Ovi
    Jul 11, 2013 at 2:35
  • $\begingroup$ @Ovi: Thinking does help. :-) It occurred to me that you might have been thinking of logarithmic differentiation of $\left(\frac{\pi}2-\tan^{-1}x\right)^{1/x}$ and have forgotten that there’s be a log in the numerator. $\endgroup$ Jul 11, 2013 at 2:40
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With $y:=\arctan\frac1x$ the limit becomes $\lim_{y\to0}y^{\tan y}=\left(\lim_{y\to0}y^y\right)^{\lim_{y\to0}\frac{\tan y}{y}}=1^1=1$.

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