All right, a little… - QuestionCove
OpenStudy (chillout):

All right, a little bit rusty in limits...

1 year ago
OpenStudy (chillout):

$$lim_{x->\infty} (x^{n}-e^{x})\,, n \in \mathbb N$$

1 year ago
OpenStudy (chillout):

I do know I'd have to use l'hospital's rule, but I'm having trouble to rewrite this as a quotient.

1 year ago
OpenStudy (chillout):

I also tried to rewrite "e" as a limit, but that didn't work out.

1 year ago
OpenStudy (freckles):

$\lim_{x \rightarrow \infty} \frac{x^n-e^x}{1} \cdot \frac{x^n+e^{x}}{x^n+e^x} \\ \lim_{x \rightarrow \infty} \frac{x^{2n}-e^{2x}}{x^n+e^{x}} \\ \text{ divide top and bottom by } e^{x}$ $\lim_{x \rightarrow \infty} \frac{\frac{x^{2x}}{e^{x}}-\frac{e^{2x}}{e^{x}}}{\frac{x^n}{e^{x}}+1}$

1 year ago
OpenStudy (chillout):

I suppose I can use the rule for all quotients now?

1 year ago
OpenStudy (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

1 year ago
OpenStudy (freckles):

since n a natural number and x goes to infinity

1 year ago
OpenStudy (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.

1 year ago
OpenStudy (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

1 year ago
OpenStudy (chillout):

All right. Just doing some head work (I had to leave for a while). The answer should be $$-\infty$$, right? ($$\large\frac{0-\infty}{0+1}$$)

1 year ago
OpenStudy (freckles):

right -inf

1 year ago