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I think you might have to use Squeeze Theorem here? I've confused myself.

$$\lim_{x \to 0+}\frac{1}{x}-\frac{1}{\arctan(x)}$$

Travis Willse
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boblo
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3 Answers3

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In THIS ANSWER, I showed using a basic inequality from geometry that for $0<x<\pi/2$ the arctangent function satisfies the inequalities

$$\frac{x}{\sqrt{1+x^2}}\le \arctan(x)\le x$$

Therefore, we can write

$$0\ge \frac1x-\frac{1}{\arctan(x)}\ge \frac{1}{x}\left(1-\sqrt{1+x^2}\right)=-\frac{x}{1+\sqrt{1+x^2}}$$

whereupon using the Squeeze Theorem we obtain the desired limit.

Mark Viola
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  • I think your second set of inequalities is backwards (and the last term should have a minus sign), but otherwise, this works. We also need to appeal to the oddness of the function to get the $x\lt0$ part. – robjohn Sep 12 '19 at 07:52
  • @robjohn Indeed. Thank you Rob; I've edited accordingly. Note that the OP is only interested in the right-sided limit, which is the reason for my not including the limit from the left. And as you stated, due to the odd symmetry, the limit from the left is the same as the limit from the right. – Mark Viola Sep 12 '19 at 15:01
  • True. I had not noticed that the question only asked for $\lim\limits_{x\to0^+}$. – robjohn Sep 12 '19 at 16:26
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Hint Rewrite the function as $$\frac{\arctan x - x}{x \arctan x} .$$ Now, apply l'Hopital's Rule (twice), or just find the leading term of the Taylor series of the numerator and denominator.

Travis Willse
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1

For $0\lt x\lt\pi/2$, $\arctan(x)\lt x$, so $\lim\limits_{x\to0^+}\left(\frac1x-\frac1{\arctan(x)}\right)\le0$. Furthermore, $\sin(x)\lt x$, therefore, $$ \begin{align} \lim_{x\to0^+}\left(\frac1x-\frac1{\arctan(x)}\right) &=\lim_{x\to0^+}\left(\frac1{\tan(x)}-\frac1x\right)\\ &\ge\lim_{x\to0^+}\left(\frac1{\tan(x)}-\frac1{\sin(x)}\right)\\ &=\lim_{x\to0^+}\frac{\cos(x)-1}{\sin(x)}\\ &=\lim_{x\to0^+}\frac{-\sin(x)}{\cos(x)+1}\\[6pt] &=0 \end{align} $$ Thus, $$ \lim_{x\to0^+}\left(\frac1x-\frac1{\arctan(x)}\right)=0 $$ Since $\frac1x-\frac1{\arctan(x)}$ is odd, we have $$ \lim_{x\to0}\left(\frac1x-\frac1{\arctan(x)}\right)=0 $$

robjohn
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