Answer
$0$
Work Step by Step
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to\infty} \dfrac{2^x-3^x}{3^x+4^x}$
The given function can be re-arranged as:
$\lim\limits_{x \to\infty} \dfrac{2^x-3^x}{3^x+4^x}(\dfrac{\frac{1}{3^x}}{\frac{1}{3^x}})=\lim\limits_{x \to\infty} \dfrac{(\frac{2}{3})^x-1}{1+(\dfrac{4}{3})^x}$
This implies that
$\dfrac{0-1}{1+\infty}=0$
