Question

Help with L'Hopital's Rule
The problem I'm working on:
lim (cotx)^x
x->0+
I would really like a step by step if possible, I'm completely stumped on this problem!

Answer 1

\(\Large\bf \color{#008353}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

\[\Large\bf\sf \lim_{x\to0^+}(\cot x)^x\]So if you try to evaluate this directly, it looks like our function is approaching the indeterminate form: \(\Large\bf\sf \infty^0\)

Answer 3

Soooo we'll have to try something fancy.

Answer 4

That's where we swap e with the lim, right?

Answer 5

No, because the base is `approaching` a large number.
The exponent is `approaching` zero.

But which one is getting there faster?
We can't necessary say that it's approaching a number to the zero power.

Answer 6

Yah plugging in a small small number is a useful :) Usually works.
But let's do the L'Hop method as the problem asks.

Yes, we have to do a sneaky little trick involving e.
\[\Large\bf\sf e^{\ln x}\quad=\quad x\]Does that look familiar?

Answer 7

Yes, so far I have:
\[e^\lim xlncotx \]
as x approaches 0+

Answer 8

Ok good.
Let's ignore the e for right now, and just pay attention to what's going on up there.\[\Large\bf\sf \lim_{x\to0^+} x \ln(\cot x)\]

Answer 9

We have to do another weird little trick from here :)

Answer 10

that's an I.F. lim as x->0+ ln(cotx)/(1/x)

Answer 11

i.f?

Answer 12

Indeterminate Form

Answer 13

True.
But it's now the I.F that we want, right?

With the way we have it written now, the function is approaching \(\Large\bf\sf \dfrac{\infty}{\infty}\)

Answer 14

So we can apply L'Hop to it from this form!

Answer 15

Okay, first L'Hop!
(csc^2x/cotx)/(-1/x^2)

Answer 16

This is where I start getting really lost

Answer 17

Negative sign on the top somewhere also yes?
(cot x)' = -csc^2x I think.

Answer 18

oops! You're right!

Answer 19

Hmm yah things are getting tricky here aren't they?
We can convert everything to sines and cosines.
It might help.

Answer 20

I'm ending up with this limit now,\[\Large\bf\sf \lim_{x\to0^+}\frac{x^2}{\sin x \cos x}\]I went through it quickly though, possible error. So try it and see if you get the same thing.

Answer 21

Let's see. . .1/cotx = tanx, or sinx/cosx and -csc^2x = 1/sin^2x, so we have (x^2)\sinxcosx YES!

Answer 22

Oops, I forgot a negative in the 1\sin^2x

Answer 23

yah it cancel's with the negative from x^2. No big deal.

Answer 24

Rewrite the denominator using Sine Double Angle Identity (will make it easier to differentiate).

Answer 25

\[\Large\bf\sf \lim_{x\to0^+}\frac{x^2}{\frac{1}{2}\sin(2x)}\]

Answer 26

And from here we can see that our function is approaching the indeterminate form 0/0.
Ok good! Still a good form that we can work with.

Time for L'Hop #2!!!

Answer 27

Trying to figure out where that 1/2 came from? XD lol

Answer 28

Yes! I'm terrible with the double angles and half angles!

Answer 29

Here is our Double Angle Identity,\[\Large\bf\sf \color{royalblue}{\sin(2x)=2\sin x \cos x}\]If we divide both sides by 2,\[\Large\bf\sf \frac{1}{2}\sin(2x)=\sin x \cos x\]The sinx cosx is the part we're replacing in our integral.

Answer 30

Okay, I see it now!

Answer 31

Let's see if I can differentiate. . .

Answer 32

2x/[1/2 cos(2x) 2] => 2x\cos(2x)

Answer 33

Ok good.

Answer 34

Are we still getting an indeterminate form as x->0 ?

Answer 35

No!, now we have 0/1

Answer 36

Ooo interesting!
So our exponent ( the whole thing we had up inside of the e ) = 0.

Answer 37

So our initial guess was correct! :) lol

Answer 38

AHA! e^0 = 1!

Answer 39

Thanks so much! You really are awesome!

Answer 40

:)

Answer 41

Np \c:/