How do I find the following limit of $f'(a+(1/x))$ without assumption of continuity of $f '$?

Question

We are given $a$ is a real number such that $f(a)=5$, $f'(a)=2$. Calculate the limit: $$\lim_{n \to \infty}\left(\frac{f\left(a+1/n\right)}{f(a)}\right)^n$$ Here is what I tried: I just raised $e$ to the power of all of that and then took $\log$ and I got that I'd need to calculate the limit of $$\exp\left( n \cdot \log\left(\frac{f(a+(1/n))}{f(a)}\right)\right).$$

Now using L'Hospitals rule, I got: $$\lim_{x \to \infty}\frac{\log(f(a+(1/x))-\log(f(a))}{1/x}=\lim_{x \to \infty}(f'(a+(1/x))/f(a+(1/x)). $$

Now $f(a+(1/x))$ approaches $f(a)=5$, because $f$ is continuous at $a$. But what does $f'((a+(1/x))$ approach? If we knew $f'$ was continuous, the answer would be $2$, and then the total answer would be $e^{2/5}$ which is the real answer, but how can you get it without knowing $f'$ is continuous at $a$?

Answer 1

Let $g(x)=\ln f(x)$. Then $g$ is differentiable at $a$ and $g'(a)=\frac{f'(a)}{f(a)}=\frac 25$. On the other hand, by definition of $g'(a)$, we have $$g'(a)=\lim_{n\to\infty}\frac{g(a+\tfrac1n)-g(a)}{\frac1n} =\lim_{n\to\infty}n(\ln f(a+\tfrac 1n)-f(a))=\lim_{n\to\infty}\ln\left(\left(\frac{f(a+\tfrac 1n)}{f(a)}\right)^n\right)$$ Finally, by continuity of $\exp$, $$\exp g'(a)=\exp \lim_{n\to\infty}\ln\left(\left(\frac{f(a+\tfrac 1n)}{f(a)}\right)^n\right)=\lim_{n\to\infty}\left(\frac{f(a+\tfrac 1n)}{f(a)}\right)^n$$

Answer 2

You can write, for $f(a)>0$, $$ f\Bigl(a+\frac{1}{n}\Bigr)=f(a)+\frac{f'(a)}{n}+o(1/n)= f(a)\Bigl(1+\frac{f'(a)}{nf(a)}+o(1/n)\Bigr) $$ so that $$ \log\Bigl(f\Bigl(a+\frac{1}{n}\Bigr)\Bigr)= \log f(a)+\log\Bigl(1+\frac{f'(a)}{nf(a)}+o(1/n)\Bigr)= \log f(a)+\frac{f'(a)}{nf(a)}+o(1/n) $$ so your limit becomes $$ \lim_{n\to\infty}\frac{\dfrac{f'(a)}{nf(a)}+o(1/n)}{\dfrac{1}{n}}=\frac{f'(a)}{f(a)} $$