Question

limit question

Answer 1

ok so i dont really get the idea of when...

Answer 2

but its negative infinity....

Answer 3

i thought dividing by infinity gives you zero so ... idk

Answer 4

As \(\Large\rm x\to-\infty\) we have \(\Large\rm \frac{e^{-x}}{x}\to\frac{\infty}{-\infty}\) yah?

Answer 5

yes

Answer 6

So we have an indeterminate form.
We want to know whether the numerator is getting to infinity faster or slower than the bottom.

Answer 7

Are you allowed to use L'Hopital's Rule for this problem?
That approach would make the most sense.

Answer 8

ummmm could you first explain to me what this means though "faster or slower than the bottom." and, ididnt learn L'hopitals rule or maybe i did but i dont know the name of it? idk

Answer 9

Think of it like a race. We want to know who is winning, the top or the bottom.

If the top is winning, we can think of it like \(\Large\rm \dfrac{\infty}{c}\) where the top is blowing up much faster. So, overall, the limit is blowing up.

If the bottom is winning, we can think of this form as something like \(\Large\rm \dfrac{c}{\infty}\). Which, as you mentioned, is clearly o overall.

Answer 10

clearly \(\Large\rm 0\) overall*

Answer 11

So which one is winning? who is blowing up faster?
The exponential, or the linear x?

Answer 12

the exponential im thinking..

Answer 13

Good good good.
The function \(\Large\rm e^{-x}\) is growing much much faster than \(\Large\rm x\).
So the top is winning, it's approaching "infinity" much much faster

Answer 14

You might learn about L'Hopital's Rule later on.
When you have something of the form:\[\Large\rm \lim_{x\to-\infty}\frac{f(x)}{g(x)}\]And it's approaching an indeterminate form of \(\Large\rm \dfrac{\infty}{\infty}\) or \(\Large\rm \dfrac{0}{0}\)
Then,

Answer 15

\[\Large\rm \lim_{x\to-\infty}\frac{f(x)}{g(x)}=\lim_{x\to-\infty}\frac{f'(x)}{g'(x)}\]

Answer 16

You can take derivative when it's in that specific form, which is actually what we had happening in this problem. Maybe you haven't learned that method yet though :o

Answer 17

ok i got every thing up to the part where you had the f prime over g prime

Answer 18

and derivative...

Answer 19

For this problem, we were approaching the indeterminate form \(\Large\rm \dfrac{\infty}{-\infty}\).

Since we have the correct form, we can apply L'Hopital's Rule and take the derivative of the top and bottom seperately.

Answer 20

\[\Large\rm L'Hop\implies\lim_{x\to-\infty}\frac{-e^{-x}}{1}\]The derivatives are ok, yes?

Answer 21

I took the derivative of the top and bottom SEPARATELY, no quotient rule or anything like that :)

Answer 22

so the numerator would be neglected in this case?

Answer 23

neglected? :o

Answer 24

oh god i mean no ummm, becuase the answer was negative infinity so you took the bottom and topseperatley and since the denominator is negative infinty soo

Answer 25

This\[\Large\rm \lim_{x\to-\infty}\frac{e^{-x}}{x}\]didn't tell us anything.
So we applied L'Hopital's rule and got this,\[\Large\rm \lim_{x\to-\infty}\frac{-e^{-x}}{1}\]Now if we "Plug in infinity", we can see that we no longer have an indeterminate form.\[\Large\rm \lim_{x\to-\infty}\frac{-e^{-x}}{1}=\frac{-e^{-(\infty)}}{1}\]

Answer 26

Woops sorry I plugged that in wrong hehe

Answer 27

\[\Large\rm \lim_{x\to-\infty}\frac{-e^{-x}}{1}=\frac{-e^{-(-\infty)}}{1}\]

Answer 28

oh i know where im confused ! how did you get like i know you said top and bottom seperatley but i just dont see it in this equation