Question

More limits....
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Answer 1

already used L'Hospital's and got \(\dfrac{2x}{e^x}\), but forgot how to do as limit approaches infinity

Answer 2

Hey there, you should apply the L'Hospital's rule again!

Answer 3

the exponential grows faster than ANY polynomial

Answer 4

Because it is still infinity/infinity.

Answer 5

l'hopital will give it, but \(e^x\) grows faster than \(x^n\) for any \(n\)

Answer 6

\(\lim_{x \to \infty }\dfrac{2x}{e^x} \to \frac{\infty}{\infty}\), crank the wheel one more time on L'Hopital's just like @steve816 is sayin

Answer 7

I have to use L'Hospital's rule twice?

Answer 8

yup

Answer 9

How do I tell when I need to use it once vs. when I need to use it twice?

Answer 10

Kinda, this is undefined, but no amount of L'Hopital will save you:

\[\lim_{x \to 0} \frac{1}{x}\]

It's specifically when you have something of an indeterminate form.

Answer 11

Ohh yes, indeterminate form, that was the word I was looking for.

Answer 12

convince yourself that \[\lim_{x\to \infty}\frac{c^n}{e^x}=0\] for any \(n\) by visualizing using hopital \(n\) itmes

Answer 13

Okay, now I'm confused and really sorry I missed this lesson. What is an indeterminate form?

Answer 14

typo there \[\large \lim_{x\to \infty}\frac{x^n}{e^x}=0\]

Answer 15

If you're not allowed to use `L'Hospital's rule`, then you can approximate e^x using a `taylor polynomial`

\[\Large e^x \approx 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3\]

the series goes on forever, but we only need up to a cubic.

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So,

\[\Large \frac{x^2}{e^x} \approx \frac{x^2}{1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3}\]

From there, divide every term by x^3 and you'll have 1/x in the numerator and the denominator will be some constant. The "1/x" will approach 0 as x approaches infinity.

Answer 16

Here are the list of indeterminate forms