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Math Help - Without using L'Hospital's Rule

  1. #1
    Junior Member BayernMunich's Avatar
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    Without using L'Hospital's Rule

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