Given $f(x)$ such that $f'(0)=2;f(0)=1$ use the maclaurin expansion of $f(x)$ to calculate $\lim_{x\to 0}(f(x))^{1/x}$.
Note: It's not given that $f(x)$ is differntiable around $x=0$
I tried using the expansion $f(x)=1+2x+R_1(x)$ but then when plugin in for the limit I need to calculate the derivative of $R_1(x)$ and I don't know anything about that. What am I missing?
EDIT: If I do this: $\lim_{x\to 0}((1+2x+R_1(x))^{1/x})=\lim_{x\to 0} (1+2x)^{1/x}$ I get $e^2$ but I'm not sure how to justify that. If it is at all correct.
We have that $f(0)=1$ and $f'(0)=2$ implies $$0=\lim_{x\to 0}\left(\frac{f(x)-1}{x-0}-2\right)=\lim_{x\to 0}\frac{f(x)-(1+2x)}{x}$$ that is $f(x)=1+2x+o(x)$. Hence, as $x\to 0$, $$f(x)^{1/x}=\exp(1/x\cdot\ln(f(x)))=\exp(1/x\cdot\ln(1+2x+o(x)))= \exp\left(\frac{2x+o(x)}{x}\right)\to e^2$$ where we used the expansion of the logarithm $\ln(1+t)=t+o(t)$ (which follow from the fact that the derivative of $\ln(1+t)$ at $0$ is equal to $1$).
P.S. I guess you are trying to evaluate the limit of $(f(x))^{1/x}$ as $x\to 0$ (not $x\to \infty)$. Note that $1+2x$ and $e^{2x}$ satisfy the given conditions but the desired limits at $\infty$ are different: $\lim_{x\to \infty}(1+2x)^{1/x}=1$ and $\lim_{x\to \infty}(e^{2x})^{1/x}=e^2$.