Proof: limit (eh – 1)/h = 1

Let t=\frac{1}{(e^h-1}\text{, so }\frac1t=e^h-1\text{, so }1+\frac1t=e^h\text{, so }h=\ln{1+\frac1t}

We seek to find the value of \displaystyle\lim_{h \to \infty}\frac{(e^h-1)}h. Using the above relationships, we get,

=\displaystyle\lim_{t \to \infty}\frac{(\frac1t)}{\ln{(1+(\frac{1}{t}))}}
=\displaystyle\lim_{t \to \infty}\frac{1}{t\ln{(1+(\frac{1}{t}))}}
=\displaystyle\lim_{t \to \infty}\frac{1}{\ln{(1+(\frac{1}{t}))^t}}

By definition,

e=\displaystyle\lim_{t \to \infty}\ln{(1+\frac1t)^t}

Therefore,

=\frac{1}{\ln{e}}

But \ln{e} is the power to which e must be raised to get e. And that power is 1 so,

\displaystyle\lim_{h \to \infty}\frac{(e^h-1)}h=1

This proof was taken from following website:

http://2000clicks.com/mathhelp/CalculusLimitExponential.aspx