How can I compute $$\lim_{u\to 0}\frac{(u+1)^\tau-1}{u}$$ without l'Hopital ($\tau>0$) ?
I wrote it as
$$\lim_{u\to 0}\frac{(u+1)^\tau-1}{u}=\lim_{u\to 0}\frac{e^{\tau\frac{\ln(u+1)}{u}u}-1}{u},$$ and tried to us that $$\lim_{u\to 0}\frac{\ln(1+u)}{u}=1,$$
but I can't conclude.
On
For $\tau >0$ an integer, apply the Binomial Theorem: $$(u+1)^{\tau} =\sum_{k=0}^{\tau}\binom{\tau}{k}u^{k}=1+\sum_{k=1}^{\tau}\binom{\tau}{k}u^{k}$$ from which we easily obtain that the limit you seek above is equivalent to finding $$\lim_{u \to 0}\left[\sum_{k=1}^{\tau}\binom{\tau}{k}u^{k-1}\right]=\binom{\tau}{1}\cdot 1 = \tau\text{.}$$ For $\tau > 0$ not an integer, rewrite $$f(u) = (u+1)^{\tau}=e^{\ln[(u+1)^{\tau}]}=e^{\tau \ln(u+1)}\text{.}$$ Assuming that you have that the derivative of $e^x$ is $e^x$, as well as the chain rule, you can see that $$f^{\prime}(u)=e^{\tau \ln(u+1)}\left[\dfrac{\tau}{u+1} \right] = e^{\ln[(u+1)^{\tau}]}\left[\dfrac{\tau}{u+1} \right]$$ and since $e$, $\ln$ are inverses, we have $$f^{\prime}(u)=(u+1)^{\tau}\left[\dfrac{\tau}{u+1} \right]$$ from which we obtain $$f^{\prime}(0)=\tau\text{.}$$
Note that $(u+1)^\tau=\exp(\tau\log(u+1))$; if you set $\tau\log(u+1)=v$, then $$ (u+1)^\tau=e^v $$ and $$ u=e^{v/\tau}-1 $$ so your limit becomes $$ \lim_{v\to0}\frac{e^v-1}{e^{v/\tau}-1}= \tau\lim_{v\to0}\frac{e^v-1}{v}\frac{v/\tau}{e^{v/\tau}-1} $$ If you consider $\lim_{v\to0}\frac{e^v-1}{v}=1$ as a known limit, you have what you're looking for.
The derivative of $f(x)=x^\tau$ can be more easily computed with the chain rule: $$ f(x)=e^{\tau\log x} $$ so $$ f'(x)=e^{\tau\log x}\cdot\frac{\tau}{x}=\tau x^\tau x^{-1}=\tau x^{\tau-1} $$