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Calculus Level 1

Detemine :
lim n e n n 2 \large \displaystyle\lim_{n\rightarrow \infty} \dfrac{ e^n }{n^2} .

Details : e is an exponential function .


1 \infty e 2 0

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2 solutions

Brock Brown
Sep 28, 2015

By L'Hopital's Rule , we can say...

L = lim n e n n 2 = L = \lim_{n \to \infty} \frac{e^n}{\color{#D61F06}{n^2}} = \frac{\infty}{\infty} \implies

L = lim n e n 2 n = L = \lim_{n \to \infty} \frac{e^n}{\color{#D61F06}{2n}} = \frac{\infty}{\infty} \implies

L = lim n e n 2 = 2 L = \lim_{n \to \infty} \frac{e^n}{\color{#D61F06}{2}} = \frac{\infty}{2} \implies

L = L = \boxed{\infty}

Useful tip. Try letting m = 1/n, so you don't have to deal with infinities.

Hint: Show that e^x is approximately equals to x for small x (when it's close to 0).

By the way, it is a common convention to hate on Lhopital rule because it is too cheesy of a solution. It's like going into a classroom and deciding to hate the 1st person you meet. In this case, L'hopital is that student.

Pi Han Goh - 5 years, 8 months ago
Otto Bretscher
Sep 12, 2015

e n > n 3 3 ! e^n>\frac{n^3}{3!} for positive n n , by Taylor Series, so e n n 2 > n 6 \frac{e^n}{n^2}>\frac{n}{6} and lim n e n n 2 = \lim_{n\to\infty}\frac{e^n}{n^2}=\boxed{\infty}

Sir can we say something like this :

e = 1 = 0 \frac { e^{ \infty } }{ \infty \infty } =\frac { 1 }{ \infty } =0

Syed Baqir - 5 years, 9 months ago

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Wrong. it is infinity/infinty = undefined.

Pi Han Goh - 5 years, 9 months ago

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Hmm, Thanks for reply :)

Syed Baqir - 5 years, 9 months ago

I also thought

Turbulent Payel - 5 years, 7 months ago

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