How to evaluate $\lim _{x\to \infty }x\left(\arctan x-\frac{\pi }{2}e^{1/x}\right)$?

Question

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 $$\lim _{x\to \infty }x\left(\arctan x-\frac{\pi }{2}e^{1/x}\right)$$

Answer 1

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?

Answer 2

With l'Hospital

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

Answer 3

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.

Answer 4

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.

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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.