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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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    Why does it seem obvious?2012-03-18

7 Answers 7

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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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\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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    @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.2012-03-20
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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. $




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    Dear David Mitra. Please help me how to draw a beautiful graph like you did?2012-03-20
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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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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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    Using L 'Hopital's is perfectly fine, as long as the asker knows of the rule. This is just my opinion.2016-08-16
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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.