ChillOut
  • ChillOut
All right, a little bit rusty in limits...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ChillOut
  • ChillOut
\(lim_{x->\infty} (x^{n}-e^{x})\,, n \in \mathbb N\)
ChillOut
  • ChillOut
I do know I'd have to use l'hospital's rule, but I'm having trouble to rewrite this as a quotient.
ChillOut
  • ChillOut
I also tried to rewrite "e" as a limit, but that didn't work out.

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freckles
  • freckles
\[\lim_{x \rightarrow \infty} \frac{x^n-e^x}{1} \cdot \frac{x^n+e^{x}}{x^n+e^x} \\ \lim_{x \rightarrow \infty} \frac{x^{2n}-e^{2x}}{x^n+e^{x}} \\ \text{ divide top and bottom by } e^{x}\] \[\lim_{x \rightarrow \infty} \frac{\frac{x^{2x}}{e^{x}}-\frac{e^{2x}}{e^{x}}}{\frac{x^n}{e^{x}}+1}\]
ChillOut
  • ChillOut
I suppose I can use the rule for all quotients now?
freckles
  • freckles
you can but e^x is getting to bigger faster than x^n or x^(2n) but you can use that rule to prove it you would just after to take the derivative infinitely many times... You will see the polynomial degree is decreasing while the e^x bottom remains the same which means the x^(2n)/e^x or the x^n/e^x approaches 0
freckles
  • freckles
since n a natural number and x goes to infinity
ChillOut
  • ChillOut
All right, thanks! It's been some years since I took limits :) I'm going to finish it on paper and will post later.
freckles
  • freckles
using l'hospital I think you should end up with \[\lim_{x \rightarrow \infty} \frac{(2n)!}{e^{x}} \text{ and } \lim_{x \rightarrow \infty} \frac{n!}{e^x}\] and then the limits are easy to evaluate
ChillOut
  • ChillOut
All right. Just doing some head work (I had to leave for a while). The answer should be \(-\infty\), right? (\(\large\frac{0-\infty}{0+1}\))
freckles
  • freckles
right -inf

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