Why here we don't use l'hopital rule?

Question

lim (root x/(1+e^x))

x-> infinity

The answer of this problem is 0 and I can't understand why we don't use l'hopital rule in here. (because, I think root x goes infinity and 1+e^x also goes infinity so it's indeterminate form.) What's the reason of this?

Answer 1

After applying L'Hospital's Rule (indeed the original is an indeterminate form), the result is:

1/2√x / (e^x) = 1/[2√x (e^x)]

lim_x->∞ = 1/∞ = 0

desmos.com/calculator/n9iobyddkp

Answer 2

Good question. You could try L'Hopital's rule here, but it isn't technically necessary.

Another method to achieve the solution is by considering growth rates. The graph of x^1/2, for all values of x greater than 1, will always grow more slowly than (1 + e^x). You can easily confirm this by taking the derivative of the two expressions separately. You'll find that x^1/2 actually grows more and more slowly as x approaches infinity, while (1 + e^x) grows at an increasing rate of e^x.

Since the growth rate of the denominator is significantly higher and increasing, and since (1 + e^x) > x^1/2 for all values of x greater than 1, we can confidently say (thinking of the right-end behavior of the graph) that the denominator is larger than the numerator and will, in fact, infinitely expand its "lead." Hence, the denominator will eventually become arbitrarily large while the growth rate of the numerator approaches zero, and it follows that the limit of the entire expression is zero.

I hope this helps!