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How do I solve this limit without using L'Hopital's rule, if at all possible?

$$ \lim_{h\to0}\frac{1 - 2^{-h}}{h}$$

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Have you defined the constant $e$? If so, how? –  JavaMan Jan 4 '13 at 21:46

3 Answers 3

up vote 1 down vote accepted

Note that $$\lim\limits_{x\to{0}}{\left( 1+x \right)^{\tfrac{1}{x}}}=e$$ is equivalent (by continuity of $\log{}$ function) to $$\lim\limits_{x\to{0}}{\dfrac{\log_a(1+x)}{x}}=\log_a{e}=\dfrac{1}{\ln{a}}.\;\;(a>0,\; a\ne{1}) \tag{*}$$ Substitute $t=\log_a(1+x),$ then $1+x=a^t, \;\; x=a^t-1$ and rewrite $({}^*)$ as $$\lim\limits_{t\to{0}}{\dfrac{t}{a^t-1}}=\dfrac{1}{\ln{a}}$$ or $$\lim\limits_{t\to{0}}{\dfrac{a^t-1}{t}}=\ln{a}.$$

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This one was most clear to me. Thank you. –  Ataraxia Feb 5 '13 at 0:16

Observe $$\lim_{h\to 0}\frac{1-2^{-h}}{h}=-\lim_{h\to 0}\frac{2^{-h}-2^{-0}}{h}=...$$ (Hint: Define the derivative of $2^{-x}$ at $0$)

EDIT: The definition of the derivative of $2^{-x}$ at $0$ is $$\lim_{h\to 0}\frac{2^{-h}-2^{-0}}{h}$$ But that is $(2^{-x})^{\prime}(0)=(-\ln 2\cdot 2^{-x})(0)=-\ln 2\cdot 2^0=-\ln 2$

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I'm not seeing it :( I see how you did that, but what do you do from there to get ln(2)? –  Ataraxia Jan 4 '13 at 22:00
    
He/She 's just saying that this is, by definition, the (negative of) the derivative of the function $2^{-x}$ at the point $0$. If $f(x) =2^{-x}$, then you're looking for $-f'(0)$. So if you know how to find this derivative, you're done. –  Brusko651 Jan 4 '13 at 22:02

You can expand the Numerator as a power series.

Using $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \dots$,

$$2^{-h} = e^{\log(2) \cdot (-h)} = 1 - \log(2)\cdot h + \frac{(\log(2)\cdot h)^2}{2} \dots$$

Therefore $$\lim_{h\to0}\frac{1 - 2^{-h}}{h} = \lim_{h\to0}\frac{\log(2)\cdot h - \frac{(\log(2)\cdot h)^2}{2} \dots}{h} = \lim_{h\to0} \left( \log(2) - \frac{(\log(2))^2h}{2} \dots \right) = \log(2)$$

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