Evaluate $\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$

Question

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?

Answer 1

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 $$\lim_{x\rightarrow0}\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}=e^1.$$

Answer 2

$$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]=\lim_{x\to0}x \cdot\dfrac{\left(\lim_{x\to0}\left(1+3x\right)^{1/3x}\right)^3}{\left(\lim_{x\to0}\left(1+2x\right)^{1/2x}\right)^2} =\lim_{x\to0}x \cdot\dfrac{e^3}{e^2}=?$$