Question

Use L'Hospital's Rule to show that "Limit as X approaches infinity that (1+1/x)*x=e

I dont know how to solve this...

Answer 1

\[\lim_{X \rightarrow \infty}\left( 1+ 1/x \right)^x= e\] This is how the equation looks like. Its just I dont know how to take the derivative of this.

Answer 2

Hmm so the first step is to make sure that it's of one of the specific indeterminate forms that we're allowed to apply L'Hospotal's to.

Looks like when x heads towards infinity, our function is heading towards,\[\large 1^\infty\]While that is of indeterminate form, we need the form,\[\large\frac{0}{0} \quad \text{or}\quad \frac{\infty}{\infty}\]

Another thing that might have helped to notice is that we can't take a derivative while there is a variable in the exponent AND base. So there are 2 reasons why we have to change the form around before we can differentiate.

Answer 3

Okay yes I got \[1^\infty \] I just couldn't figure out how to take the derivative of the equation. Once I get that I know to just put infinity back in and see if it matches to one of the indeterminate forms.

Answer 4

This next step can be a little confusing. Do you remember back to this identity?\[\large e^{\ln x}=x\]This is due to the fact that the exponential and the logarithm are inverse functions of one another.

We're going to be applying this rule in reverse.
We're STARTING with something involving x, and we're going to rewrite it as e^ln x.

Answer 5

\[\huge \lim_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x \quad \rightarrow \quad \lim_{x \rightarrow \infty}e^{\ln\left(\left(1+\frac{1}{x}\right)^x\right)}\]

Answer 6

Understand how we did that? :O It's a little tricky.

Answer 7

Yes, I do get that. :) Thank you! I forgot about the whole e^in of x=x so now that makes sense :) So now I just put infinity in for X and that should be my answer correct?

Answer 8

I think we'll still have a problem, from here we wan to use rules of logarithms to move that X out of the log.

Then we can get it into one of the 2 indeterminate forms that we want, and we'll be allowed to use L'Hop from there.

Answer 9

From here, we just want to be working WITHIN the exponent.
I know this is a little weird, but try to imagine yer floating in the sky, and the base e is below you.

We'll just be dealing with this part of the problem,\[\huge \lim_{x \rightarrow \infty}{\ln\left(\left(1+\frac{1}{x}\right)^x\right)}\]

Answer 10

Ohh okay! Lol I love the way you explain it btw :P

Answer 11

Recall we can apply this rule,\[\large \log(b^a) \quad \rightarrow \quad a\cdot \log(b)\]Which will give us, Oh umm.. I'mma drop the limit for now, I don't want to keep writing it.\[\huge x\cdot \ln\left(1+\frac{1}{x}\right)\]

Answer 12

Grrr gimme 2 minutes to finish my sammich >:O it's getting cold.. I knew I shouldn't had started a problem right now lol

Sec brb :D

Answer 13

Lol no problem :) Thanks :D

Answer 14

Once we are done with this I actually have a couple other problems and if you wouldnt mind helping me with them, I would be very thankful :)

Answer 15

If we were to let x approach infinity at this point, you would see that we have a problem approaching something like this,\[\large 0\cdot \ln(1)\quad \rightarrow \quad 0\cdot 0\]Hmm that's still no good, we need the QUOTIENT of functions to be able to apply L'Hop.

Answer 16

Ahh that's frustrating, couldn't reconnect D:< Finally!

Answer 17

So we will take the quotient of 1/X?

Answer 18

This next step is a little tricky, we're gonna do something silly with the x term. We can rewrite it as the reciprocal of the reciprocal. We're sort of like.... flipping it, twice... which doesn't change the value of it. Lemme know if this is confusing or not.\[\huge x\cdot \ln\left(1+\frac{1}{x}\right) \quad \rightarrow \quad \frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}\]

Answer 19

Ahh yer too fast >:O lol

Answer 20

Okay....Ohh! I see! Then we would put infinity in and get something like \[\infty/infty\]

Answer 21

Yah :) In this case, since x is approaching infinity, we get 0/0 which is also an acceptable form for L'Hop.

Answer 22

Yay! Thank you!

Answer 23

So apply L'Hop from this point, you'll get a nice cancellation.
And then DON'T FORGET ABOUT THE E. Remember? We're floating in da sky! :O

Answer 24

Okay what would I do with the E?

Answer 25

Ignore it until the very very end.
Do some L'Hop. Figure out what you're getting for a limit. Then as a final step, write it as e^ to that value.

Answer 26

So I take the deriavtive of \[\ln \left( 1+1/x \right)/\left( 1/x \right)\]

Answer 27

yah

Answer 28

Well remember that you're taking the derivative of the top and bottom SEPARATELY. So you don't have an ugly quotient rule problem like it might look at first glance.

Answer 29

Lol yeah thats what I thought at first but I remembered I was supposed to look at them SEPARATELY.

Answer 30

Okay so how would I take the derivate of \[\ln \left( 1+1/x \right)\] I have the derrivate of 1/x which is just -1/x^2

Answer 31

\[\huge \left(\ln x\right)'=\frac{1}{x}\]\[\huge \left(\ln\left(1+\frac{1}{x}\right)\right)'\quad =\quad \frac{1}{\left(1+\frac{1}{x}\right)}\left(1+\frac{1}{x}\right)'\]

Answer 32

Don't forget the chain rule! :O That's what the prime term is on the right, have to differentiate that part still. Should give you something similar to the derivative of 1/x that you already took earlier ;U

Answer 33

So now I put in infinity and I get 1? So its e^1?

Answer 34

Yah looks good! :)
Remember at the start what they were asking for?
They actually gave you the answer, they said PROVE IT!! NOW!!!

Answer 35

Lol wow they did :P Lol well thank you :D