Q:

Find $\lim _{x\to \infty \:}\left(\frac{2^x-3^x}{3^x+4^x}\right)$

$$\lim _{x\to \infty \:}\left(\frac{2^x-3^x}{3^x+4^x}\right)$$
When substituting, it has the form 
$$\frac {\infty-\infty}{\infty+\infty}$$
I think it has to do with L'Hopital rule, But I couldn't put in an acceptable form $\dfrac{\infty}{\infty}$ or $\dfrac{ 0}0$
I think also, since the numerator is a difference and the denominator is addition, the limit will go to zero?  Is this "rough way" useful? 

A:

The general principle that I recommend is to divide top and bottom by the dominant term of the denominator.   In this case, $4^x$:
$$\frac{2^x-3^x}{3^x+4^x}=\frac{(\frac{2}{4})^x-(\frac{3}{4})^x}{(\frac{3}{4})^x-1}$$
Now, $\lim_{x\to\infty}(\frac{2}{4})^x=0$, $\lim_{x\to\infty}(\frac{3}{4})^x=0$ so the limit is $$\frac{0-0}{0-1}=0$$

A:

$$\lim_{x\to \infty} \frac{2^x-3^x}{3^x+4^x}$$
$$=\lim_{x\to \infty} \frac{(2/3)^x-1}{1+(4/3)^x}$$
$$=\frac{-1}{\infty}$$
$$=0$$

A:

The best way to do this is to factor out $3^x$ from the numerator, and $4^x$ from the denominator.  Then $\bigg(\dfrac{3}{4}\bigg)^x \rightarrow 0$, and the rest of the fraction goes to $-1.$

