Question

How do I begin to take the limit of (4x)^(1/x)?

Answer 1

Limit as x approaches what value?

Answer 2

Oh infinity. sorry about that.

Answer 3

So our function is approaching the `indeterminate form`: \(\Large\sf \infty^{0}\)

Answer 4

So we need to try and do something fancy to see what's going on.
It's not in the correct form to apply L'Hopital just yet.
Maybe we can fix that though.

Answer 5

Here is a handy trick we can use:\[\Large\sf x=e^{\ln x}\]

Answer 6

\[\Large\sf \lim_{x\to\infty}4x^{1/x}\]So we'll apply this idea to our problem,
both exponentiate and take the log,\[\huge\sf e^{\ln(\lim_{x\to\infty}4x^{1/x})}\]We can pull the limit operator out of the log,\[\huge\sf e^{\lim_{x\to\infty}\ln(4x^{1/x})}\]Let's ignore the e base for a moment and just pay attention to what's going on up in the exponent.\[\Large\sf \lim_{x\to\infty}\ln(4x^{1/4})\]

Answer 7

Woops typo on that last line*\[\Large\sf \lim_{x\to\infty}\ln(4x^{1/x})\]

Answer 8

Rules of exponents allows us to bring the 1/x outside of the log, yes?

Answer 9

\[\Large\sf \lim_{x\to\infty}\frac{1}{x}\ln(4x)\]

Answer 10

To be able to apply L'Hopital, we want something of the form:\[\Large\sf \lim_{x\to\infty}\frac{f(x)}{g(x)}=\frac{0}{0},\qquad or\qquad \lim_{x\to\infty}\frac{f(x)}{g(x)}=\frac{\infty}{\infty}\]

Answer 11

Ok I'll stop talking for a sec :) What you thinking?

Answer 12

Okay. Yeah there is a bit of fancy footwork going on there. I'm not awfully familiar with infinity to the zero problems so I'm just looking over what you've got so far.

Answer 13

So you put the whole limit inside of the natural log when you do e to the ln?

Answer 14

Yah I'm applying this idea,\[\Large\sf \color{orangered}{x}=e^{\ln \color{orangered}{x}}\]To our problem,\[\Large\sf \color{orangered}{\lim_{x\to\infty}4x^{1/x}}=e^{\ln(\color{orangered}{\lim_{x\to\infty}4x^{1/x}})}\]

Answer 15

Okay I got you. So next you would use L'hopitals rule on
\[\frac{ \ln 4x }{ x }\] ?

Answer 16

Oh the 4 is in brackets with the x?
Ah I've been a little sloppy with my notation, sorry about that.
Shouldn't change anything though.

Answer 17

\[\Large\sf \lim_{x\to\infty}\frac{\ln(4x)}{x}\]is giving us the indeterminate form: \(\Large\sf \dfrac{\infty}{\infty}\)

Ah yes, good!
Looks like we can apply L'Hop from here!

Answer 18

Yeah so it would end up being
\[\frac{ 4 }{ x }\]
and then
\[\frac{ 0 }{ 1 }\] right?

Answer 19

\[\Large\sf \lim_{x\to\infty}\frac{\ln(4x)}{x} \qquad L'H\to \qquad \lim_{x\to\infty}\frac{4}{x}\]This limit is approaching 4/infty,
which does not allow us to apply L'Hop a second time.
This is no longer an indeterminate form.

Answer 20

Is that what you were trying to do maybe?

Answer 21

Really it was a sequence that I was trying to find the limit of. So in that case it diverges?

Answer 22

No.
We have a fraction.
The numerator is constant, the denominator is getting really really big.

So as a whole, the fraction is approaching zero, right?\[\Large\sf \frac{4}{2bajillion}\approx 0\]

Answer 23

It turns out that by applying L'Hop a second time, you would have simply gotten lucky.
But it wasn't correct to do so, so you gotta be careful! :)

Answer 24

Hahah alright. I usually make that mistake. Thanks a bunch!

Answer 25

But also recall this only takes care of the exponent part of our problem.
We still have the base e, right?

Answer 26

That's what I was thinking. Apparently the limit is actually 1 when I throw it into wolfram alpha. so e to the 0 is 1?

Answer 27

Because we took the e out?

Answer 28

Yes good, so the whole limit mess simplified to zero.
We're left with e^0.
Which simplifies to 1, yes.

Answer 29

Awesome Okay thanks for your help!

Answer 30

np!