Evaluate limit $\lim_{x\to0+} \frac{(1+2x)^x-1}{(1+3x)-1}$

Question

how do you solve

$$\lim_{x\to0+} \frac{(1+2x)^x-1}{(1+3x)-1}$$

• method 1 using l'Hospital's rule

• method 2 using definition of $e$

  $$e=\lim_{x\to 0}(1+x)^{1/x}$$

  thanks!

Answer 1

Here's what I would do $$(1+2x)^x-1=e^{x\ln (1+2x)}-1$$ $$=\frac{e^{x\ln (1+2x)}-1}{x\ln (1+2x)}x\ln (1+2x)$$

Now $$\frac{e^{x\ln (1+2x)}-1}{x\ln (1+2x)}\to 1$$ by the limit $$\lim\limits_{x\to 0}\frac{e^x-1}{x}=1$$

so thus your question becomes

$$\frac{x\ln (1+2x)}{x\ln (1+3x)}=\frac{\ln (1+2x)}{x}\frac{x}{\ln (1+3x)}\to \frac{2}{3}$$