## University Calculus: Early Transcendentals (3rd Edition)

$\infty$
Consider $f(x)=\lim\limits_{x \to 0} (\dfrac{1}{x^4}-\dfrac{1}{x^2})=\lim\limits_{x \to 0}\dfrac{1-x^2}{x^4}=\lim\limits_{x \to 0}\dfrac{1-x^2}{x^4}$ Let $a(x)=\lim\limits_{x \to 0}1 -x^2=1$ Then $\lim\limits_{x \to 0}\dfrac{1-x^2}{x^4}=\lim\limits_{x \to 0}\dfrac{1}{x^4}=\dfrac{1}{0}=\infty$