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.

  • $\begingroup$ @PedroM.: Because I want to prove that $(x^\tau)'=\tau x^{\tau-1}$, therefore, I can't use it. $\endgroup$ – user380364 Dec 26 '17 at 14:37
  • $\begingroup$ Bernoulli's Inequality? $\endgroup$ – user123641 Jan 9 '18 at 3:13

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 $$ f'(x)=e^{\tau\log x}\cdot\frac{\tau}{x}=\tau x^\tau x^{-1}=\tau x^{\tau-1} $$

  • $\begingroup$ That a perfect answer. Thanks a lot. $\endgroup$ – user380364 Dec 26 '17 at 14:52

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 $$f^{\prime}(0)=\tau\text{.}$$

  • $\begingroup$ It's what I want to prove (i.e. $(x^\tau)'=\tau x^{\tau-1}$) therefore, I can't use this result. $\endgroup$ – user380364 Dec 26 '17 at 14:36
  • 1
    $\begingroup$ @PedroM. This is NOT using L-Hospital, but the definition of the derivative. $\endgroup$ – Clarinetist Dec 26 '17 at 14:39
  • $\begingroup$ using the definition of the derivative and l'Hopital are equivalent ;-) $\endgroup$ – user380364 Dec 26 '17 at 14:39
  • $\begingroup$ @user380364 Okay, since you're not allowed to use the power rule... let me edit this. $\endgroup$ – Clarinetist Dec 26 '17 at 14:41
  • $\begingroup$ @user380364 Actually, one thing that you need to note is that you can't prove this for general $\tau$ without using implicit differentiation or logarithmic differentation. Are you allowed to use these? For $\tau$ an integer, the proof can be done without these. $\endgroup$ – Clarinetist Dec 26 '17 at 14:43

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