Easy difference quotient question

Sheneron
Messages
360
Reaction score
0

Homework Statement


Using difference quotient I am trying to find f '(0) for 2^x. Basically my question is a questions of algebra but I will show you what I have done thus far.

the limit is as x -> 0

\frac{2^x - 2^0}{x - 0}

\frac{2^x - 1}{x}

So my question is what can I do to get x off the bottom.
 
Physics news on Phys.org
You've probably already proved that the limit x->0 of (e^x-1)/x=1. Try to use that.
 
Thanks, that helped.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top