$\lim \frac{10+3^x}{20-3^x}$ 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. 

Notice, $$\lim_{x\to \infty}\frac{10+3^x}{20+3^x}$$
$$=\lim_{x\to \infty}\frac{3^x\left(\frac{10}{3^x}+1\right)}{3^x\left(\frac{20}{3^x}-1\right)}$$
$$=\lim_{x\to \infty}\frac{\frac{10}{3^x}+1}{\frac{20}{3^x}-1}$$
$$=\frac{0+1}{0-1}=-1$$
