Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 71

$0$

Here, $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to\infty} \dfrac{2^x-3^x}{3^x+4^x}$ This implies that $\lim\limits_{x \to\infty} \dfrac{2^x-3^x}{3^x+4^x}(\dfrac{1/3^x}{1/3^x})=\lim\limits_{x \to\infty} \dfrac{(2/3)^x-1}{1+(4/3)^x}$ Thus, $\dfrac{0-1}{1+\infty}=0$