Find limit as $t$ approaches $0$

Question

Suppose $f(x)$ is differentiable at $x_0$ and $f(x_0)=1$. Find $\lim_{t\to0}f(x_0+t)^\frac{1}{t}$.

I tried using the epsilon delta definitions but I'm quite stuck.

Answer 1

HINT

Note that

$$f(x_0+t)^\frac{1}{t} =e^{\frac{\log(f(x_0+t))}{t}}=e^{\frac{\log(f(x_0+t))-\log f(x_0)}{t-0}}$$

then note that exponential is continuos and evaluate

$$\lim_{t\to0} \frac{\log(f(x_0+t))-\log f(x_0)}{t-0} $$