Answer
$\lim\limits_{x \to 0}\frac{x^3-5x^2}{x^2}=-5
$
Work Step by Step
$\lim\limits_{x \to 0}\frac{x^3-5x^2}{x^2}\\
divide\,both\,numerator\,and\,denominator\,\,by\,\,x^2 \\
\lim\limits_{x \to 0}\frac{x^3-5x^2}{x^2}=\lim\limits_{x \to 0}\frac{\frac{x^3-5x^2}{x^2}}{\frac{x^2}{x^2}}\\
=\lim\limits_{x \to 0}\frac{x-5}{1}=\lim\limits_{x \to 0}x-5=-5 \\
$
