Find $\lim_{x \to 0} (f(3+1/x)/f(3))^x$ Let $f(3)=5$ and $f'(3)=2$. Find $$\lim_{x\rightarrow0} \left(\frac{f(3+\frac{1}{x})}{f(3)} \right)^{x}$$
My question is :


*

*Is the problem correct?

*Is the answer $e^{\frac{2}{5}}$ correct?

*If the answers of above questions are yes, Can I apply L'Hôpital's rule to the problem?

*Is there any faster solution?

 A: I assume the limit is needed when $x \to \infty$ and not when $x \to 0$.
If $f$ admits a Taylor series expansion around $3$, then $$f(3+1/x) = f(3) + f'(3)/x + \mathcal{O}(1/x^2)$$
Hence, $$\dfrac{f(3+1/x)}{f(3)} = 1 + \dfrac{f'(3)/f(3)}{x} + \mathcal{O}(1/x^2)$$
$$\left(\dfrac{f(3+1/x)}{f(3)}\right)^x = \left(1 + \dfrac{f'(3)/f(3)}{x} + \mathcal{O}(1/x^2) \right)^x$$
Recall that $$\lim_{x \to \infty} \left(1+ \dfrac{a}x + o(1/x) \right)^x = e^a$$
A: Take a logarithm and write $g = \log \circ f$. You get
$$
\lim_{x \to 0} x(g(3 + \frac{1}{x}) - g(3))
$$
Now if $x$ were going to $\infty$, then setting $h = 1/x$, this would be (by definition of the derivative!) $g'(3)$ which is $\frac{f'(3)}{f(3)} = \frac{2}{5}$ so that the final answer is indeed $e^{2/5}$. So maybe there is a typo in the problem?
A: You want to compute 
$$
\lim_{x \to 0} \frac{f(3+ \frac 1x)^x}{f(3)^x}
$$
But the denominator goes to $1$ as $x \to 0$. Therefore you can use the fact that limits of quotients is the quotient of limits and not worry about using things such as the Hospital's rule. But notice that if the limit exists,
$$
\lim_{x \to -\infty} f(3+x)^{1/x} = \lim_{x \to 0} f(3 + \frac 1x)^x = \lim_{x \to \infty} f(3+x)^{1/x}
$$
so the limit has nothing to do with the function's behavior at $y=3$. Did you have any particular function $f$ in mind?
