Without using L'Hospital's Rule, how does one show that lim - infinity of (X^2)/e^x = 0?

Question

Without using L'Hospital's Rule, how does one show that lim -> infinity of (X^2)/e^x = 0?

anonymous · Answer

That's supposed to read X-> infinity ...

nenadmatematika · Answer

function e^x grows much faster than x^2 when x goes to infinity, so when x^2 reaches some big number, we can consider that e^x ''reaches'' infinity....so if you put big number over infinity the result iz zero, actually limit is zero....also if you would like to know...$$\ln x <<x^k (k \in R) << k^x << x! <<x^x$$ this means that every limit when x goes to infinity of quotient of these functions, going from left to the right is zero, and in opposite direction is infinity....it's a very useful idea....if you are confused, $$e^x \approx2.7$$ so you can treat him as any number k^x.....I hope I was clear...good luck :D

cristiann · Answer

Consider the limit lim_{n→∞}((n²)/(eⁿ));
let b=e-1>0
C_{n}^{k} is a notation for "n choose k" -- the binomial coefficient from Newton binomial
0<((n²)/(eⁿ))=((n²)/((b+1)ⁿ))=((n²)/(∑_{k=0}ⁿC_{n}^{k}b^{k}))<((n²)/(C_{n}³b³))=((3!n²)/(n(n-1)(n-2)b³))→0 as n→∞
    and use for example "The Squeeze Theorem" to get lim_{n→∞}((n²)/(eⁿ))=0
    This line of reasoning works for any n^{α} with appropriate modifications.

cristiann · Answer

just a file to see the writing better