How can I solve this limit without L'Hopital's rule? [closed]

Normally, this is done using "notable limits". For instance, $$ \lim_{x\to 0}\frac{1-\cos x}{x} =0 , \quad \lim_{x\to 0} \frac{\ln (x+1)}{x} = 1. $$

So,

$$ \lim_{x\to 0} \frac{1-\cos x}{\ln(x+1)} = \lim_{x\to 0}\left(\frac{1-\cos x}{x}\cdot \frac{x}{\ln(x+1)}\right) = 0 \times 1 = 0 $$

As the user Clayton said right we need to use here the Taylor series.

It is very well known, that $$ \bbox[lightgreen] { \cos(x)=\left(1-\frac{x^2}{4}+\frac{x^4}{24}-....\right) }, \\ \bbox[lightblue] { \ln(1+x)=\left(x-\frac{x^2}{2}+\frac{x^3}{3}-....\right) }, \implies \\ \implies{\lim_{x\to0}\frac{1-\cos(x)}{\ln(1+x)}}=\lim_{x\to0}\frac{\left(\frac{x^2}{4}-\frac{x^4}{24}+....\right)}{\left(x-\frac{x^2}{2}+\frac{x^3}{3}-....\right)}= \\ =\lim_{x\to0}\frac{x\left(\frac{x}{4}-\frac{x^3}{24}+....\right)}{x\left(1-\frac{x}{2}+\frac{x^2}{3}-....\right)}=\lim_{x\to0}\frac{0}{1}=0, \implies \\ \implies \bbox[pink] { {\lim_{x\to0}\frac{1-\cos(x)}{\ln(1+x)}}=0 }. $$