# How do you find lim 1/x^2-1/x as x->0?

Jan 8, 2017

${\lim}_{x \to 0} \frac{1}{{x}^{2}} - \frac{1}{x} = + \infty$

#### Explanation:

You can simplify the function in this way:

$\frac{1}{{x}^{2}} - \frac{1}{x} = \frac{1}{x} ^ 2 \left(1 - x\right)$

Now:

${\lim}_{x \to 0} \frac{1}{{x}^{2}} = + \infty$

${\lim}_{x \to 0} \left(1 - x\right) = 1$

So:

${\lim}_{x \to 0} \frac{1}{{x}^{2}} - \frac{1}{x} = + \infty$

graph{1/x^2-1/x [-10, 10, -5, 5]}