# Lim x->0 tan^-1x/x solve?

Jan 23, 2018

$1$

#### Explanation:

${\lim}_{x \to 0} {\tan}^{- 1} \frac{x}{x} \to \frac{0}{0}$

So we can try L'Hopital's rule:

For the numerator:

$\frac{d}{\mathrm{dx}} {\tan}^{-} 1 x = \frac{1}{1 + {x}^{2}}$

and the denominator:

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

There fore we get:

${\lim}_{x \to 0} {\tan}^{- 1} \frac{x}{x} = {\lim}_{x \to 0} \frac{\frac{1}{1 + {x}^{2}}}{1} = \frac{\frac{1}{1}}{1} = 1$

Jan 23, 2018

Alternate way of thinking of it, but $\underline{\text{less rigorous}}$

${\lim}_{x \to 0} \frac{{\tan}^{- 1} \left(x\right)}{x} = 1$

#### Explanation:

One thing to consider is approximations of trigonometric functions when $x$ is particulaly small

The first thing to note is that color(red)(tanx approx x  for $x$ being small

This is due to $\cos x \approx 1$ and $\sin x = x$ for $x$ being small

Hence $\tan x = \sin \frac{x}{\cos} x \approx \frac{x}{1} \approx x$

So hence ${\tan}^{- 1} x$ is simply $\tan x$ reflected in the line $y = x$

But we know the inverse of $y = x$ is just $y = x$

$\implies {\tan}^{- 1} \left(x\right) \approx x$ for small $x$

$\implies {\tan}^{- 1} \frac{x}{x} \approx \frac{x}{x} \approx 1$ when $x$ is very small

${\lim}_{x \to o} \frac{{\tan}^{- 1} x}{x} = {\lim}_{x \to 0} 1$

$= 1$

We can say this as $x$ gets small, to be particular, $x \to 0$ but we color(blue)(underline ("couldn't") do this if we had the limit of something like $x \to \frac{1}{10}$

Like I said at the beggining this method is a very non- rigorous approach, and doesnt always work on all functions

I'd advise you to use $L ' H$ rule for this problem, as answered prior