How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But it didn't work.
Please Help me.
 A: Without using L'Hopital's rule or Taylor series:
$\begin{align} \lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}&=\lim_{x \to 0}\frac{ \frac{\tan^{-1} x}{x}}{\frac{1-\cos x}{x^2}} \; \text{(Divided Numerator and Denominator by $x^2$)}\\ 
&=\frac{\lim\limits_{x \to 0} \frac{\tan^{-1} x}{x}}{\lim\limits_{x \to 0}\frac{1-\cos x}{x^2}}\\
&=\frac{1}{1 \over 2}=2 \end{align}$
$\begin{align}\because \lim_{x \to 0} \frac{\tan^{-1} x}{x}&=\lim_{y \to 0} \frac{y}{\tan y} \;\;(\text{substituted}\; y=\tan^{-1} x, \;y\to 0\text{ as}\; x \to 0)\\
&=\frac1{\lim\limits_{y \to 0} \frac{\tan y}{y}}\\
&= \frac{1}{1}=1 \;\quad \left(\because \lim\limits_{y \to 0} \frac{\tan y}{y} =1\right)\end{align} $
And,
$\begin{align}\lim\limits_{x \to 0}\frac{1-\cos x}{x^2}&=\lim\limits_{x \to 0}\frac {2\sin^2\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2}\\&={1\over 2}\left(\lim\limits_{\frac{x}{2}\to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2\\&={1\over 2}(1)^2 \quad \; \left(\because \lim\limits_{y\to 0}\frac{\sin y}{y}=1\right)\\&={1\over 2}\end{align}$
A: $\tan y \approx y + \frac{y^3}{3}$ from the Taylor series. We can discard the $y^3/3$ term to get $\arctan x \approx x$, as we only need the $x^2$ term and below.
Then:
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x} = \lim_{x \to 0} \frac{x^2}{1-\cos x} \frac{1 + \cos x}{1 + \cos x} = 2.$$
A: You idea is fine indeed standard limits suffices and we don't need l'Hospital's rule nor Taylor's series.
The problem you faced after the substitution is for the $\cos(\tan y)$ which stopped the resolution.
To overcome with that we can proceed in two ways:
1) Separate the limit in 2 factors
This way corresponds to what indicated by Aman Kushwaha's answer, that is
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}=\lim_{x \to 0} \frac{x^2 }{1-\cos x}\frac{\tan^{-1} x}{x}=\lim_{x \to 0} \frac{x^2 }{1-\cos x}\cdot \lim_{x \to 0} \frac{\tan^{-1} x}{x}$$
using the substitution $x=\tan y$ only for the second factor.
2) Proceed further form you last step
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }=\dots$$
as follows
$$\dots=\frac{\tan^2y }{1-\cos(\tan y) }\frac{  y}{\tan y }=\frac{\tan^2y }{1-\cos(\tan y) }\frac{  y}{\sin y }\cos y\to 2\cdot 1\cdot 1=2$$
A: You're in a $\dfrac00$ situation, so try Hopital.
A: Hint. From $\cos (A+B)=\cos A\cos B-\sin A\sin B$ with $B=A$ we have $\cos 2A=\cos^2A-\sin^2A=(1-\sin^2A)-\sin^2 A=1-2\sin^2 A.$ Hence $2\sin^2 A=1-\cos 2A.$  With $A=2x$ therefore $$2\sin^2(x/2)=1-\cos x.$$ So if $\cos x\ne 1$ then $$\frac {x\tan^{-1}x}{1-\cos x}=\frac {x\tan^{-1}x}{2\sin^2(x/2)}=$$ $$=\left(\frac {(x/2)}{\sin (x/2)}\right)^2\cdot 2\cdot \frac {\tan^{-1}x}{x}.$$ Remark: The "half-angle formulae" are usually presented as $$\frac {1-\cos x}{2}=\sin^2(x/2)$$ and $$\frac {1+\cos x}{2}=\cos^2(x/2).$$
A: Surprised no one explicitly used L'Hopital's rule.
$\displaystyle \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2}.$
Derivative of the numerator is therefore
$\displaystyle \tan^{-1}(x) + \frac{x}{1 + x^2}.$
Derivative of the denominator is $\sin(x)$
Since both derivatives evaluate as $(0)$ at $(x = 0)$, L'Hopital's rule must be re-applied.
Next derivative of numerator is
$\displaystyle \frac{1}{1 + x^2} + \frac{1 + x^2 - 2x^2}{\left(1 + x^2\right)^2}.$
At $(x=0)$, this evaluates to $(2)$.
Next derivative of denominator is $\cos(x)$.
At $(x = 0)$, this evaluates to $(1)$.
Therefore, the limit is $\displaystyle \frac{2}{1} = 2.$
