$\displaystyle \lim_{x\rightarrow0}\frac{-42(ln(1+x))+42x}{x^2(e^x+4)}=$

$\displaystyle \frac{-42[x-\frac{x^2}{2}+\frac{x^3}{3}...]+42x}{x^2[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+4]}=$

$\displaystyle \frac{[21x^2-14x^3...]}{[5x^2+x^3+\frac{x^4}{2!}...]}$

Division with $\displaystyle x^2$ gives

$\displaystyle \frac{21-14x...}{5+x+\frac{x^2}{2!}...}=$ and when$\displaystyle x\rightarrow0$ we get

$\displaystyle \frac{21-0}{5+0+0}=\frac{21}{5}$??