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I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used (without L'Hôpital if is possible)? Thanks $$\lim _{x\to \infty }x\left(\arctan x-\frac{\pi }{2}e^{1/x}\right)$$

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  • $\begingroup$ It is a good idea to rewrite this as $\lim_{y \to 0^+} \frac{\arctan(1/y) - \frac{\pi}{2} e^y}{y} = \lim_{y \to 0^+} \frac{\operatorname{arccot}(y) - \frac{\pi}{2} e^y}{y}$. Does this resemble anything familiar? $\endgroup$ – Ian Apr 25 '16 at 14:29
  • $\begingroup$ You can try at least squeeze theorem limit, $arctanx$ is bounded. $\endgroup$ – Camilo Acevedo. Apr 25 '16 at 14:56
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Hint. One may use $$ \arctan x= \frac{\pi}2-\arctan \frac1x,\quad x>0, $$ and, by the Taylor series expansions, as $x \to \infty$, $$ \begin{align} \arctan \frac1x&=\frac1x+O\left(\frac1{x^2} \right)\\\\ e^{\large \frac1x}-1&=\frac1x+O\left(\frac1{x^2} \right). \end{align} $$ Can you take it from here?

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Lets put $t = 1/x$ so that $t \to 0^{+}$ as $x \to \infty$. We have \begin{align} L &= \lim_{x \to \infty}x\left(\arctan x - \frac{\pi}{2}e^{1/x}\right)\notag\\ &= \lim_{t \to 0^{+}}\dfrac{\arctan(1/t) - \dfrac{\pi}{2}e^{t}}{t}\notag\\ &= \lim_{t \to 0^{+}}\dfrac{\arctan(1/t) - \dfrac{\pi}{2} + \dfrac{\pi}{2}(1 - e^{t})}{t}\notag\\ &= \lim_{t \to 0^{+}}\dfrac{-\arctan t + \dfrac{\pi}{2}(1 - e^{t})}{t}\notag\\ &= -\lim_{t \to 0^{+}}\left(\frac{\arctan t}{t} + \frac{\pi}{2}\cdot\frac{e^{t} - 1}{t}\right)\notag\\ &= -\left(1 + \frac{\pi}{2}\right)\notag \end{align} For most of the limit evaluations a combination of "algebra of limits", "standard limits" and "squeeze theorem" suffices and only significantly difficult problems necessitate the use of powerful tools like L'Hospital's Rule and Taylor series expansions.

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With l'Hospital

$$\lim_{x\to\infty}\frac{\arctan x-\frac\pi2e^{1/x}}{\frac1x}=\lim_{x\to\infty}\frac{\frac1{1+x^2}+\frac\pi{2x^2}e^{1/x}}{-\frac1{x^2}}=-\lim_{x\to\infty}\left(\frac{x^2}{x^2+1}+\frac\pi2e^{1/x}\right)=-1-\frac\pi2$$

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With $x=\dfrac1t$,$$\frac{\dfrac\pi2-\arctan(t)-\dfrac\pi2e^t}t=-\frac\pi2\frac{e^t-e^0}t-\frac{\arctan(t)-\arctan(0)}t.$$

The solution follows, either by recognizing known limits, or recognizing the definition of derivatives (hence first degree Taylor coefficient), which amounts to L'Hospital.


If you want to avoid derivatives by all means, $$\frac{e^t-1}t=\lim_{n\to\infty}\frac{\left(1+\frac1n\right)^{nt}-1}t=\lim_{n\to\infty}\frac{\left(1+\frac tn\right)^{n}-1}t.$$

By the binomial theorem, this expression under the limit is

$$1+\frac{n-1}{2n}t+\frac{(n-1)(n-2)}{3!n^2}t^2+\frac{(n-1)(n-2)(n-3)}{4!n^3}t^3+\cdots<\frac1{1-t}$$ and this converges to $1$ for $t\to0$.

Then, by a change of variable, we turn

$$\frac{\arctan(t)}t$$ into $$\frac t{\tan(t)}.$$

Considering the trigonometric circle, inscribe a rectangle triangle in a sector of amplitude $t$. The area of the triangle is comprised between the areas of two sectors (or radii $\cos(t)$ and $1$), and ignoring a factor $\dfrac12$

$$t\cos^2(t)<\sin(t)\cos(t)<t$$ so that

$$1<\frac{\tan(t)}t<\frac1{\cos^2(t)}.$$

By squeezing you get the desired result.

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