[College Calc 01-03. Limits of Functions] #01

Note that

$\lim_{x\to 0}\frac{x^2}{e^x-x} = \frac{0}{1 - 0} = 0;$

be careful not to use l'hopital on the limit above. As for the second term,

$\lim_{x\to 0}\frac{6^x-e^x}{x} = \lim_{x\to 0}\frac{(6^x-1)-(e^x-1)}{x}=\ln 6 - \ln e = -1 + \ln 6.$

Therefore, $p=-1$ and $q=6$ such that $p+q = 5.$

We can use L'hopital when we combine the fractions! $\lim_{x\rightarrow} \left(\frac{x^2}{e^x-x}+\frac{6^x-e^x}{x}\right)=\lim_{x\rightarrow 0} \frac{x^3+6^xe^x-e^{2x}-x6^x+xe^x}{xe^x-x^2}\overset{\textrm{L'H} \frac{0}{0}}{=}\lim_{x\rightarrow 0} \frac{3x^2+\ln (6e)6^xe^x-2e^{2x}-6^x-\ln 6 x6^x+e^x+xe^x}{e^x+xe^x-2x}=\ln 6+1-2-1+1=\boxed{\ln 6-1}$