The limit as x->0^+ (from above zero) of $\displaystyle ln(x+1)/x$ is 1, I know this from punching numbers into my calculator (and from Maple). My question is how can i show this algebraically?
Hey kingsolomonsgrave.
For this problem you get a 0/0 case which is where L'Hopitals rule comes in (and since you are approaching it from the right and not the left you can use this: if you approached it from the left then it wouldn't exist altogether in the real numbers).
d/dx[ln(x+1)] = 1/(x+1) and d/dx(x) = 1. This will give a limit that can be evaluated.
If you are not familiar with L'Hopitals rule it is basically a rule that you can use if you get an indeterminate form like 0/0 or infinity/infinity type limits.
Hello, kingsolomonsgrave!
Do you know this definition of $\displaystyle e$?
. . $\displaystyle \lim_{u\to\infty}\left(1 + \tfrac{1}{u}\right)^u \;=\;e$
$\displaystyle \text{Show that: }\:\lim_{x\to0^+}\,\frac{\ln(x+1)}{x} \:=\:1$
We have: .$\displaystyle \lim_{x\to0^+}\tfrac{1}{x}\ln(x+1) \;=\;\lim_{x\to0}\ln(1+x)^{\frac{1}{x}} $
Let $\displaystyle u =\tfrac{1}{x} \quad\Rightarrow\quad x \,=\,\tfrac{1}{u}$
Note that: if $\displaystyle x\to 0$, then $\displaystyle u \to \infty.$
$\displaystyle \text{So we have: }\:\lim_{u\to\infty}\left[\ln\left(1 + \tfrac{1}{u}\right)^u\right] \;=\;\ln\underbrace{\left[\lim_{u\to\infty}\left(1 + \tfrac{1}{u}\right)^u\right]}_{\text{This is }e} \;=\;\ln(e) \;=\;1$