Answer
$-1$
Work Step by Step
Here, $\lim\limits_{x \to -\infty} f(x)=\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}$
This implies that
$\lim\limits_{x \to -\infty} \dfrac{2^x+4^x}{5^x-2^x}(\dfrac{1/2^x}{1/2^x})=\lim\limits_{x \to -\infty} \dfrac{1+(4/2)^x}{(5/2)^x-1}$
Thus, $\dfrac{1+0}{0-1}=-1$