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anonymous

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    That's supposed to read X-> infinity ...

  2. nenadmatematika
    • 4 years ago
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    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

  3. cristiann
    • 4 years ago
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    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.

  4. cristiann
    • 4 years ago
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    just a file to see the writing better

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