Prove that $\lim_{x\rightarrow a}(1+u(x))^{v(x)}=e^{\lim_{x\rightarrow a}u(x)\cdot v(x)}$ if $u(x)\rightarrow 0$ and $v(x)\rightarrow \infty$

Question

My study book gives out the following equation as a rule, but without any further explanation. I don't know where to start, as my knowledge of limits and exponentials is quite limited. $$(u(x)\rightarrow 0) \wedge (v(x) \rightarrow \infty) \Rightarrow \lim_{x\rightarrow a}(1+u(x))^{v(x)}=e^{\lim_{x\rightarrow a}u(x)\cdot v(x)}$$ I'm interested in its proof, so any help would be appreciated. Thanks in advance.

Answer 1

Write $$ (1+u(x))^{v(x)}=\left[(1+u(x))^{1/u(x)}\right]^{u(x)v(x)}. $$ Then, \begin{gather} \lim_{x\to a}(1+u(x))^{1/u(x)}=\lim_{z\to 0}(1+z)^{1/z}=e,\\\lim_{x\to a}u(x)v(x)\text{ exists (the question assumes this implicitly)} \end{gather} and the result you seek follows.

Answer 2

We don't need the hypothesis $$\lim_{x\to a}v(x)=+\infty.$$

we only need

$$\lim_{x\to a}v(x)\left(\ln(1+u(x))-u(x)\right)=0$$

to have

$$e^{v(x)\ln(1+u(x))}\sim e^{u(x)v(x)}\;(x\to a)$$