Limit using hopital rule

Question

lim as x tends to infinity (e^-x²) (x^3+2x^2+1) 

Answer 1

Hi Alex,

 

We want to solve

 

lim_{x →∞) exp(-x²)(x³ + 2x² + 1)

 

= lim_{x →∞) (x³ + 2x² + 1)/exp(x²) → ∞/∞, which is an indeterminate form. So, we must now employ l'Hopital's rule.

 

For such indeterminate cases, l'Hopital's rule states that the lim f(x)/g(x) = lim f'(x)/g'(x) where the prime denotes differentiation with respect to x.

 

In the last form of the limit I wrote above, let's take the derivative of the numerator and the denominator. We get

 

lim_{x →∞) (3x² + 4x)/[2xexp(x²)] = lim_{x →∞) (3x + 4)/[2exp(x²)] → ∞/∞

 

If at first you don't succeed, try again. L'Hopital's rule can be applied multiple times until you get a result. So, let's apply it again to the last form of the limit that we had above. Let's calculate the derivative of the numerator and denominator and take the limit again:

 

lim_{x →∞) 3/[4xexp(x²)] → 3/∞ → 0