$\lim \frac{10+3^x}{20-3^x}$ as $x$ goes to $\infty$ 
Find $\lim \frac{10+3^x}{20-3^x}$ as $x$ goes to $\infty$

I tried $\displaystyle \lim_{x \to \infty} \frac{10+3^x}{20-3^x} =\lim_{x \to \infty} \frac{\frac{d}{dx}10+3^x}{\frac{d}{dx}20-3^x}= -\lim_{x \to \infty}\frac{3^x \log{3}}{3^x \log{3}} = -1.$
However, this is wrong. 
EDIT: It's in fact correct. I do wish to see if there are other alternatives, though. 

Factor $3^x$ in numerator and denominator.
