Limit problem as x goes to infinity

Question

$\lim_{x\to \infty } e^x-x^2$ This is of the form $\infty - \infty$.
Is it legal to use L'hopital's rule here to get $e^x-2x$ and again to get $e^x-2$ = $\infty$

I know that the limit is $\infty$ but not sure my approach is legal.

Answer 1

Your approach is not legit. In this case for this particular limit, you can just need to manipulate your limit to get a L'Hospitals applicable case, as :

$$\lim_{x \to \infty} (e^x - x^2) = \lim_{x \to \infty} e^x\left(1-\frac{x^2}{e^x}\right) = \lim_{x \to \infty}e^x \cdot \lim_{x \to \infty}\bigg(1-\frac{x^2}{e^x}\bigg)$$

Now, you can apply L'Hospital's Rule in the fraction and proceed to showing that the limit equals $\infty$.

Note that in any case, for large $x$, it's also $e^x > x^2$ in a more non-rigorous approach, as the exponential term wins out on the square of $x$.

Answer 2

Write your term in the form $$e^x-x^2=e^x\left(1-\frac{x^2}{e^x}\right)$$