Prove $\lim_{x\to 0^+}(e^x-1+x)^\frac 2x=0$ Prove $\lim_{x\to 0^+}(e^x-1+x)^\frac 2x=0$
The limit is $0$ when ${x\to 0^+}$ and $\infty$ when ${x\to 0^-}$
I have tried using logaritms and $\lim_{x\to a}f(x)^{g(x)}=e^{\lim_{x\to a}(f(x)-1)(g(x))}$, but  it doesn´t work.
 A: Whenever you see a limit with an exponent like that, you should strongly consider taking the logarithm of it (and usually applying L'Hopital's rule when you can, but not here).
Let $y = \lim_{x \to 0} (e^x - 1 + x)^{2/x}$. Then 
$$\ln(y) = \ln(\lim_{x \to 0} (e^x - 1 + x)^{2/x}) = \lim_{x \to 0} \ln\left((e^x - 1 + x)^{2/x}\right)$$
If you're concerned about the rigorous details of why we can pass the limit through, know that it is sufficient for a function to be continuous for us to pass a limit through it.
Now,
$$\ln(y) = \lim_{x \to 0} \frac{2}{x} \cdot \ln(e^x - 1 + x)$$
Notice now that as $x \to 0^+$ (i.e., from the right), then $\ln(e^x - 1 + x)$ tends to $-\infty$, and since $x > 0$ when we appraoch from the right, we have $\frac{2}{x} \cdot \ln(e^x - 1 + x) \to -\infty \Rightarrow y \to 0$. Similarly, when we approach from the left side, we will have $\frac{2}{x} \cdot \ln(e^x - 1 + x) \to \infty$ so that $y \to \infty$.
In summary, the limit as we appraoch from the right is $0$, where as the limit diverges when you appraoch from the left.
