## Thomas' Calculus 13th Edition

$-1$
Here, we have $\lim\limits_{x \to -\infty} f(x)=\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}$ Re-arrange as: $\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}(\dfrac{\dfrac{1}{2^x}}{\dfrac{1}{2^x}})=\lim\limits_{x \to -\infty} \dfrac{1+(\dfrac{4}{2})^x}{(\dfrac{5}{2})^x-1}$ and $\dfrac{1+0}{0-1}=-1$