I was asked to find two limits.

Let $f$ be differentiable function at $x=1$ and $f(1)>0$.

  • $$\lim_{n \rightarrow \infty}\left(\frac{f\left(1+\frac{1}{n}\right )}{f(1)} \right)^{\frac{1}{n}}$$

Let $\frac{1}{n}=h$ so the limit becomes $$\lim_{h \rightarrow 0}\left(\frac{f\left(1+h \right)}{f(1)} \right)^h=\lim_{h \rightarrow 0} \left(\frac{f(1+h)-f(1)}{f(1)}+1 \right)^h$$

How may I continue?

Same for $\lim_{x \rightarrow 1} \left(\frac{f(x)}{f(1)} \right)^{\frac{1}{\log(x)}}$. I defined $h=\log(x)$ and tried to continue with no luck.

Please help, thank you!


Your approach is correct and it can be continued in the following manner. Let $$g(h) = \frac{f(1 + h) - f(1)}{f(1)}$$ so that $\lim_{h \to 0^{+}}g(h) = 0$ and also note that $\lim_{h \to 0^{+}}g(h)/h = f'(1)/f(1)$. Now we have find the limit $$L = \lim_{h \to 0^{+}}(1 + g(h))^{h}$$ Clearly if we apply logarithms we get

$\displaystyle \begin{aligned} \log L &= \lim_{h \to 0^{+}}\log\{1 + g(h)\}^{h}\\ &= \lim_{h \to 0^{+}}h\log\{1 + g(h)\}\\ &= \lim_{h \to 0^{+}}hg(h)\frac{\log\{1 + g(h)\}}{g(h)}\\ &= 0\cdot 0\cdot 1 = 0\end{aligned}$

so that the answer is $L = 1$.

The second part with $$L = \lim_{x \to 1}\left(\frac{f(x)}{f(1)}\right)^{1/\log x}$$ can be handled similarly. Let us take logs to get

$\displaystyle \begin{aligned}\log L &= \lim_{x \to 1}\frac{1}{\log x}\log\left(\frac{f(x)}{f(1)}\right)\\ &= \lim_{h \to 0}\frac{1}{\log (1 + h)}\log\left(\frac{f(1 + h)}{f(1)}\right)\\ &= \lim_{h \to 0}\frac{h}{\log (1 + h)}\frac{1}{h}\log\{1 + g(h)\}\\ &= \lim_{h \to 0}\frac{h}{\log (1 + h)}\cdot\frac{g(h)}{h}\cdot\frac{\log\{1 + g(h)\}}{g(h)}\\ &= 1\cdot\frac{f'(1)}{f(1)}\cdot 1 = f'(1)/f(1)\end{aligned}$

Hence $L = e^{f'(1)/f(1)}$

Update: In response to some comments I am providing a bit more elaboration. First note that

$\displaystyle \begin{aligned}\lim_{h \to 0}\frac{g(h)}{h} &= \lim_{h \to 0}\frac{f(1 + h) - f(1)}{hf(1)}\\ &= \lim_{h \to 0}\frac{f(1 + h) - f(1)}{h}\cdot\frac{1}{f(1)}\\ &= \frac{f'(1)}{f(1)}\end{aligned}$

Next while taking logarithms we use the logarithms to base $e$. Note that in analysis/calculus the symbol $\log$ always means a natural logarithm and logarithm to other bases have to be indicated explicitly with the base like $\log_{b}a$. Also note the fact that it is only for the natural logarithms to the base $e$ for which the following limit formula holds: $$\lim_{h \to 0}\frac{\log(1 + h)}{h} = 1$$ it does not hold if the base of the logarithm is not $e$.

In fact one can also say that the base $e$ is a "natural" choice for the base of logarithms precisely to make this limit equal to $1$. If the base were something else (like $10$ or some other positive number $a$) then the above limit would not be a simple number like $1$.

  • $\begingroup$ Great answer, thanks! $\endgroup$ – Galc127 Dec 27 '13 at 11:30
  • $\begingroup$ Why is $\lim_{h \to 0^{+}}g(h)/h = f'(1)/f(1)$ ? $\endgroup$ – GinKin Dec 28 '13 at 14:33
  • $\begingroup$ Also, $L = e^{f'(1)/f(1)}$ ? it could have been any base it doesn't have to necessarily be e right ? $\endgroup$ – GinKin Dec 28 '13 at 15:15
  • $\begingroup$ @GinKin: I have provided more explanation in my answer. Hope it helps to answer your queries. $\endgroup$ – Paramanand Singh Dec 29 '13 at 3:56
  • $\begingroup$ Thanks for the explanation. I thought that just log mean base 10 and only ln means the natural log... $\endgroup$ – GinKin Dec 29 '13 at 20:00

You can also write $f(x)=f(1)+f'(1)(x-1)+o(x-1)$ as $x \to 1$. In the second limit this leads to $$ \left[ \left( 1+ \frac{f'(1)}{f(1)}(x-1)+o(x-1) \right)^{\frac{f(1)}{f'(1)}\frac{1}{x-1}} \right]^{\frac{x-1}{\log x}} $$ and you can use $\lim_{\varepsilon \to 0} \left(1+\varepsilon \right)^{\frac{1}{\varepsilon}}$.


In your first question , just substitute h in the 2nd last step from where you will get the limit as 1.

Your second question is of the form $1^{\infty}$.

Hint: If limit of ${f(x)}^{g(x)}$ is of the form $1^{\infty}$ ,then

limit = $\lim e^{(f(x) - 1)g(x)}$

  • $\begingroup$ First, thank you for your answer. Second, I still don't get it. There is no need of using derivatives? It seems that I can just plug in h=0 in the first step after subsitute $h=\frac{1}{n}$... And I really don't understand what you wrote about the second limit. $\endgroup$ – Galc127 Dec 25 '13 at 18:24
  • $\begingroup$ @Galc127 $a^x = e^{x ln(a)}$. You must be familiar with this. $\endgroup$ – user2369284 Dec 25 '13 at 18:27

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