# Proving $\lim_{x \to \infty} (-x^3 + x^2) = - \infty$ using L'Hôpitals rule

I wanted to prove that

$$\lim_{x \to \infty} (-x^3 + x^2) = - \infty$$

which is obvious. But when I tried using L'Hôpitals rule, I ran into a wrong answer.

My try was the following:

$$f(x) = -x^3 + x^2 = x^2 (-x + 1) = \frac{-x +1}{x^{-2}}$$

Now take the derivative of the function over and under the fraction:

$$\lim_{x \to \infty} \frac{-x +1}{x^{-2}} = \lim_{x \to \infty} \frac{-1}{-2x^{-3}}$$

But

$$\lim_{x \to \infty} \frac{-1}{-2x^{-3}} = \lim_{x \to \infty} 0.5 x^3 = \infty$$

which now gives the wrong result. What am I overlooking?

• you can't apply L'hospital's rule here – haqnatural Aug 7 '17 at 15:12

You just can't apply L'Hopital rule, since $\frac{-x+1}{x^{-2}}$ is not an indeterminate form as $x\to \infty$.