> 
> 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.