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Quick question:

I came across the following limit: $$\lim_{x\rightarrow 0^{+}}\frac{\arctan(x)}{x}=1.$$ It seems like the well-known limit: $$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1.$$ Can anyone show me how to prove it?

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  • $\begingroup$ Why does it seem obvious? $\endgroup$ – Jonas Meyer Mar 18 '12 at 20:16
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We can make use of L'Hopital's rule. Since $\frac{d}{dx}\arctan x=\frac{1}{x^2+1}$ and $\frac{d}{dx}x=1$, we have $$\lim\limits_{x\to0^+}\frac{\arctan x}{x}=\lim\limits_{x\to0^+}\frac{1}{x^2+1}=1.$$

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    $\begingroup$ Just a simple remark: since the limit is the derivative of $\tan^{-1}$ at $0$, it is rather circular or unnecessary to use L'Hopital's Rule here. Look at the accepted answer. $\endgroup$ – Pedro Tamaroff♦ Mar 18 '12 at 23:13
  • $\begingroup$ @PeterT.off True. I did not realize that until it was posted. $\endgroup$ – Alex Becker Mar 18 '12 at 23:16
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    $\begingroup$ Using L 'Hopital's is perfectly fine, as long as the asker knows of the rule. This is just my opinion. $\endgroup$ – c650 Aug 16 '16 at 21:34
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$$\lim_{x\rightarrow 0^{+}}\frac{\arctan(x)}{x}= \lim_{h\rightarrow 0^{+}}\frac{\arctan(0+h) -\arctan(0)}{h} = \arctan'(0) = \frac{1}{1+0^2} = 1$$

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  • $\begingroup$ Definitely the best one. +1 $\endgroup$ – Patrick Da Silva Mar 18 '12 at 23:11
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    $\begingroup$ [...] which is essentially repeating the proof of l'Hôpital's rule. $\endgroup$ – kahen Mar 18 '12 at 23:41
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    $\begingroup$ it's an easy special case of l'Hôpital's rule. $\endgroup$ – Dustan Levenstein Mar 19 '12 at 0:07
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    $\begingroup$ @kahen: that's not even a special case of l'Hospital's rule, it's just the definition of the derivative. The hard work in l'Hospital's rule is that the existence of the limit of the quotient follows from the existence of the limit of the quotient of the derivatives. $\endgroup$ – t.b. Mar 20 '12 at 19:26
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Recall (see the diagram below) that for $0\le t<{\pi\over2}$:

$$\tag{1} \sin t \le t \le \tan t. $$ Taking $t =\arctan x$ in $(1)$, we have, for $x>0$: $$ \sin\bigl(\arctan(x)\bigr)\le \arctan(x)\le x. $$ But $$ \sin\bigl(\arctan (x)\bigr) ={x\over \sqrt{1+x^2}}; $$ whence, for $x>0$: $$ {x\over \sqrt{1+x^2}}\le \arctan(x)\le x. $$ So, for $x>0$, we have $$ {1\over \sqrt{1+x^2}}\le {\arctan(x)\over x}\le 1; $$ and it follows from the Squeeze Theorem that $$ \lim_{x\rightarrow0^+} {\arctan(x)\over x}=1. $$



enter image description here

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  • $\begingroup$ Nice diagram :-) It is pretty similar to this one. $\endgroup$ – robjohn♦ Mar 18 '12 at 22:55
  • $\begingroup$ @David Mitra: Would you please say me how did you draw this diagramm? $\endgroup$ – Américo Tavares Mar 19 '12 at 18:01
  • $\begingroup$ @AméricoTavares using the Javascript library JSXgraph. If you like, I could link to the source. $\endgroup$ – David Mitra Mar 19 '12 at 18:03
  • $\begingroup$ @DavidMitra: Yes, please. Thanks for the information. I don't know whether I am able to install it, but I could try with the help of others. $\endgroup$ – Américo Tavares Mar 19 '12 at 18:08
  • $\begingroup$ @AméricoTavares The source for JSXgraph is at the link in my previous comment. The source to my diagram is here. JSXgraph allows you to make nice interactive diagrams; in the link I gave, you can click on and drag the point $C$ around (there are a few things I still need to do in the diagram, such as having the text $A_2$ move appropriately, and fix the highlighting). Please forgive my abysmally inefficient coding skills :) $\endgroup$ – David Mitra Mar 19 '12 at 18:15
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If you don't yet have access (which is often the case) to such relatively advanced tools as derivatives, L'Hopital's rule, and series expansion, here is a very simple proof:

Once you know:

$$\lim_{x \rightarrow 0} \frac{\sin x}{x}= 1$$

You can prove that

$$\lim_{x \rightarrow 0} \frac{\tan x}{x}= 1$$

Indeed,

$$\lim_{x \rightarrow 0} \frac{\tan x}{x}= \lim_{x \rightarrow 0} \frac{\sin x}{x \cdot \cos x}= \lim_{x \rightarrow 0} \frac{\sin x}{x} \lim_{x \rightarrow 0} \frac{1}{\cos x}= 1\cdot1 = 1$$

Now you make a simple substitution:

$$t = \arctan x \implies x = \tan t$$

$$x \rightarrow 0 \implies t \rightarrow 0$$

Finally,

$$\lim_{x \rightarrow 0} \frac{\arctan x}{x} = \lim_{t \rightarrow 0} \frac{t}{\tan t} = 1$$ (the last limit equals $1$, as proved above).

If you were actually looking for the proof $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ then there are plenty of nice unit circle proofs on the internet. Maybe you could try this one.

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Since $\lim\limits_{x\to0}\arctan(x)=0$, letting $x=\tan(\theta)$ yields $$ \lim_{x\to0}\frac{\arctan(x)}{x}=\lim_{\theta \to0}\frac{\theta}{\tan(\theta)}\tag{1} $$ and $(1)$ is shown to be $\frac11$ in equation $(5)$ of this answer.

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Can you do the Taylor series' expansion of $\arctan (x)$? If you can, then it is easy to solve your limit. If you cannot, refer to this.

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If you know that $\tan x \underset{x \rightarrow 0}{\sim} x $ you could compute :

$x = \arctan(\tan x) \underset{x \rightarrow 0}{\sim} \arctan x$ and then $ \frac{\arctan x}{x} \underset{x \rightarrow 0}{\rightarrow} 1$

Edit : sorry I don't see the proof above which use the same idea.

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