# Limit of (a^x -1)/x when x approaches 0

• March 21st 2010, 03:35 PM
Bart
Limit of (a^x -1)/x when x approaches 0

This was a problem we got last semester (book of exercises). We didn't have to solve this but I'm still wondering how to solve it.
The tip is to substitute a^x - 1 by t.
I would enter the formula in LaTeX but I haven't found yet how to enter a limit in LaTeX-code.
• March 21st 2010, 03:52 PM
Miss
Quote:

Originally Posted by Bart

This was a problem we got last semester (book of exercises). We didn't have to solve this but I'm still wondering how to solve it.
The tip is to substitute a^x - 1 by t.
I would enter the formula in LaTeX but I haven't found yet how to enter a limit in LaTeX-code.

Let $f(x)=a^x$ , then $f(0)=1$.

$\lim_{x\to 0} \frac{a^x-1}{x}=\lim_{x\to 0} \frac{a^x-f(0)}{x-0}=f'(0)=a^x ln(a) |_{x=0}=ln(a)$
• March 21st 2010, 05:52 PM
Quote:

Originally Posted by Bart

This was a problem we got last semester (book of exercises). We didn't have to solve this but I'm still wondering how to solve it.
The tip is to substitute a^x - 1 by t.
I would enter the formula in LaTeX but I haven't found yet how to enter a limit in LaTeX-code.

Alternatively, in terms of you substitution

$\lim_{x\to 0}\left(\frac{a^x-1}{x}\right)$

$a^x-1=t,\ a^x=t+1$

$x=0\ \Rightarrow\ 1=t+1\ \Rightarrow\ t=0$

$lna^x=ln(t+1)$

$xlna=ln(t+1)$

$x=\frac{ln(t+1)}{lna}$

$\frac{1}{x}=\frac{lna}{ln(t+1)}$

$\lim_{x\to 0}\left(\frac{a^x-1}{x}\right)=\left(lna\right)\lim_{t\to 0}\left(\frac{t}{ln(t+1)}\right)$

Finally use

$\lim_{t\to 0}\left[ln(t+1)\right]=t$
• March 23rd 2010, 11:09 AM
Bart
Quote:

Alternatively, in terms of you substitution

$\lim_{x\to 0}\left(\frac{a^x-1}{x}\right)$

$a^x-1=t,\ a^x=t+1$

$x=0\ \Rightarrow\ 1=t+1\ \Rightarrow\ t=0$

$lna^x=ln(t+1)$

$xlna=ln(t+1)$

$x=\frac{ln(t+1)}{lna}$

$\frac{1}{x}=\frac{lna}{ln(t+1)}$

$\lim_{x\to 0}\left(\frac{a^x-1}{x}\right)=\left(lna\right)\lim_{t\to 0}\left(\frac{t}{t+1}\right)$

Finally use

$\lim_{t\to 0}\left[ln(t+1)\right]=t$

Thank you both, this was the solution I needed.
I forgot to mention that we had to solve all limits without l' Hôspital, just for the algebraic exercise.
• March 23rd 2010, 11:41 AM
Hi Bart,

I had a typo on the 2nd last line.

You could finally use the series expansion for $ln(1+x)$

$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.......$

As x approaches zero, the higher powers of x become increasingly insignificant
in comparison to x,

hence

$ln(1+t)\ \rightarrow\ t,\ as\ t\ \rightarrow\ 0$
• March 23rd 2010, 09:43 PM
Bart
I thought so.
Thanks.