I want to evaluate the following limit:
$$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$$
We have
$$\left(\frac{1+3x}{1+2x}\right)^{\frac 1 x} = e^{\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}\equiv e^{h(x)}$$
$$h(x)= {\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}$$
The argument of $\log\left(\frac{1+3x}{1+2x}\right)$ tends to 1, therefore:
$$\log\left(\frac{1+3x}{1+2x}\right)\sim\frac{1+3x}{1+2x}-1=\frac{x}{1+2x}\sim\frac {x}{2x}=\frac 1 2$$
This means that
$$h(x)\sim\frac 1 2 \cdot \frac 1 x = \frac 1 {2x}$$
We finally have for $x\rightarrow0$:
$$e^{h(x)}=e^{\frac {1}{2x}}\rightarrow+\infty$$
This should lead to an indeterminate form $[0\cdot\infty]$. Any hints on how to evaluate that limit?
Best Answer
Recall that $x\to 0$ implies $1+2x\sim 1$. Hence you should have $$\log\left(\frac{1+3x}{1+2x}\right)\sim\frac{1+3x}{1+2x}-1=\frac{x}{1+2x}\sim\frac{x}{1}=x.$$ and therefore $$\lim_{x\rightarrow0}\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}=e^1.$$