Question

Limit as x approaches infinity of (3x)^(1/3x)

step by step help please? :)

Answer 1

Introduce a logarithm.

Answer 2

huh?

Answer 3

Rather generally, if \(\lim_{x \rightarrow c}{f(x)} = L, then \lim_{x \rightarrow c}{ln(f(x))} = ln(L)\). Can you use this lovely little hint?

Answer 4

I need to take the natural log of f(x)?

Answer 5

If you want to use my little hint, that will help you get to the natural log of the original limit.

Answer 6

I'm not understanding.... ;/

Answer 7

ln((3x^(1/3x))= 3^(1/3x)x^(1/3x)

Answer 8

Okay, one clarifying question. Is the exponent \(\dfrac{1}{3x}\) or \(\dfrac{1}{3}x\)?

Answer 9

What happened to your logarithm? It was there for a moment.

Answer 10

huh?

Answer 11

Please clarify the original problem statement.

\(\lim_{x\rightarrow\infty}(3x)^{1/(3x)}\) or \(\lim_{x\rightarrow\infty}(3x)^{(1/3)x}\)

Answer 12

^ first one

Answer 13

Awesome. We need \(\lim_{x\rightarrow\infty (3x)^{1/(3x)}}\)

If we assume that the limit exists, we have \(\lim_{x\rightarrow\infty (3x)^{1/(3x)}} = L\). I just called it"L" because I don't know what it is, yet.

Using my little hint, we have: \(\lim_{x\rightarrow\infty} ln\left((3x)^{1/(3x)}\right) = ln(L)\).

Answer 14

Make sense so far?

Answer 15

yeah I understand that.. and ln((3x^((1)/(3x))= 3^((1)/(3x))x^((1)/(3x))

Answer 16

No, that's no good. Your logarithm disappeared. \(\ln\left((3x)^{1/(3x)}\right) = \dfrac{1}{3x}ln(3x) = \dfrac{ln(3x)}{3x}\). We're just using logarithm rules.

Answer 17

ohh... what rule is that specifically? ^

Answer 18

Log Rule #3 (They don't really have numbers. :-)) -- \(log\left(a^{b}\right) = b\cdot log(a)\)

Answer 19

hahah.. then why'd you go from 1/3x*ln(3x) to ln(3x)/3x?

Answer 20

That's just multiplication. \(\dfrac{1}{a}\cdot b = \dfrac{b}{a}\).

Answer 21

hahhaha.. my gosh, i feel stupid! Okay, so now we're at..
\[\lim_{x \rightarrow \infty} \frac{ \ln(3x) }{ 3x }\]

Answer 22

^ that = 0 But, that's not the answer.. the answer is 1?

Answer 23

How did you arive at the conclusion that the limit is zero?

Answer 24

factoring, using LP rule, and then solving, I got 0?

Answer 25

That doesn't quite make sense. Something still going wrong, here, What did you factor?

Answer 26

I factored out the constant. Maybe I should stop trying to go ahead and just follow lol

Answer 27

I was afraid of that. There is no constant to factor out. ln(3x) = ln(3) + ln(x). No factoring available.

Okay, enough of that. Let's consider numerator and denominator separately...

What happens to the numerator as x increases without bound in the positive direction?

What happens to the denominator as x increases without bound in the positive direction?

Answer 28

what do you mean without bound in the positive direction?

Answer 29

These mean about the same thing:

1) x increases without bound in the positive direction
2) \(x\rightarrow\infty\)
3) Some folks might say "x approaches infinity". I just don't like that.

Answer 30

I'm honestly unsure on how to answer this.. How do I figure it out?

Answer 31

If you graphed this, y = 3x, what would it look like?

Answer 32

Close engouh. Now just look at it. As x continues to the right, what will y do?

Answer 33

goes to infinity, keeps increasing.

Answer 34

Perfect. What about y = ln(3x)? x increases and y does what?

Answer 35

increases to 3 and then goes constant basically?

Answer 36

No, that's not right. It also increases without bound, just not as fast, so you might think it levels off. Really, y just keeps getting larger as x increases. Do you believe?

Answer 37

does it just slowly increase?

Answer 38

Using logs base 10

log(1) = 0
log(10) = 1
log(100) = 2
log(1000) = 3
log(10000) = 4

See, it just keeps increasing, albeit WAY slower.

Answer 39

ohh okay

Answer 40

Okay, the point of all this "increase" talk is to consider \(\dfrac{ln(3x)}{3x}\). What we just discovered, was that both numerator and denominator continue to increase. In the notation of Indetermiante Forms, this is type \(\dfrac{\infty}{\infty}\). I'm hoping really hard that you have seen this sort of thing.

Answer 41

Yeah with LP rule right?

Answer 42

Awesome. I like to spell it out, so I can laugh at how close it is to "Hospital". l'hopital's rule woudl suggest we do what with such an indeterminate form?

Answer 43

It's French ;p we take the derivative of both the denom and num.. (1/x)/(3)

Answer 44

Excellent. And what is the limit of that as \(x\rightarrow\infty\)?

Answer 45

How do I solve for that.. ? I usually put the infinity limits on a calculator, but this gave mme -3E-6...?

Answer 46

That's you calculator's way of saying "A really, REALLY small number". What non-negative number can you think of that is less than any other nonnegative number?

Answer 47

....? I'm confused..

Answer 48

The limit of that expression, with increasing x, is zero (0).

Answer 49

how?

Answer 50

How is \(\lim_limits{x\rightarrow\infty}\dfrac{1}{3x} = 0\)? Are you sure you don't want to take that one back? The numerator is constant and the denominator increases.

Answer 51

no, I get it. But that's not the answer.

Answer 52

Yes it is. You're just a little too jumpy. :-) We've been talking so long, it is no surprise that you forgot where we started.

We did NOT just say L = 0, we just said ln(L) = 0. Now, go find L.

Answer 53

No, I've written everything down that we've talked about.
and L is the f(x) in the original equation right?

Answer 54

Right, and we abandoned that right away. We've been working with ln(f(x)) all along.

Answer 55

so, now I do ln((3x)^(1/3x))=0?

Answer 56

And then get rid of the logarithms. If ln(L) = 0, the \(L = e^{0} = 1\) and we have reached the end of this one.

Answer 57

How does (3x)^(1/3x)=e^(0)?

Answer 58

What we have is a theorem.

If lim(f(x)) = L, then lim(ln(f(x)) = ln(L)
also
If lim(ln(f(x))) = ln(L), then lim(f(x)) = L

This is not an algebraic manipulation. This is the application of a theorem.