# Thread: Without using L'Hospital's Rule

1. ## Without using L'Hospital's Rule

$\displaystyle \lim_{x\to 0} \dfrac{ 2^{arctan(x)} - 2^{arcsin(x)} } { 2^{tan(x)} - 2^{sin(x)} }$

2. ## Re: Without using L'Hospital's Rule

Let me try. First let me write

$\frac{2^{\tan^{-1}x} - 2^{\sin^{-1}x} }{2^{\tan x} - 2^{\sin x} }$ $\displaystyle = \frac{ \frac{2^{\tan^{-1}x} - 1}{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{x} - \frac{2^{\sin^{-1}x} - 1}{\sin^{-1}x} \cdot \frac{\sin^{-1}x}{x} }{ \frac{2^{\tan x} - 1}{\tan x} \cdot \frac{\tan x}{x} - \frac{2^{\sin x} - 1}{\sin x} \cdot \frac{\sin x}{x} }$

so our limit is

$\lim_{x \to 0} \displaystyle \frac{ \frac{2^{\tan^{-1}x} - 1}{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{x} - \frac{2^{\sin^{-1}x} - 1}{\sin^{-1}x} \cdot \frac{\sin^{-1}x}{x} }{ \frac{2^{\tan x} - 1}{\tan x} \cdot \frac{\tan x}{x} - \frac{2^{\sin x} - 1}{\sin x} \cdot \frac{\sin x}{x} }.$

Now we can cancel some terms since the term $\frac{2^h-1}{h}$ appears is each term where $h = \{\sin x, \tan x, \tan^{-1}x, \sin^{-1}x\}$ and when $x \to 0$, these become the same.

Thus,

$\lim_{x \to 0} \displaystyle \frac{ \frac{\tan^{-1}x}{x} - \frac{\sin^{-1}x}{x} }{ \frac{\tan x}{x} - \frac{\sin x}{x} }$

(we let $x = \tan x$ and $x = \sin x$ in the first and second terms in numerator) so

$\lim_{x \to 0} \displaystyle \frac{ \frac{x}{\tan x} - \frac{x}{\sin x} }{ \frac{\tan x}{x} - \frac{\sin x}{x} }$

and simplifying and cancelling gives

$\lim_{x \to 0} \displaystyle \frac{ -x^2 }{ \sin x \tan x } = -1$

noting that all I used was $\lim_{x \to 0} \frac{\sin x}{x} = 1$.