Homework Help: A limit question

1. Oct 20, 2007

transgalactic

i added a file with the question and how i tried to solve it

lim [2+n^(1/3)]^(1/n)

in the end i always get (infinity)^0 type of object
then i got stuck
what to do next??

Attached Files:

• 6.JPG
File size:
4.5 KB
Views:
94
2. Oct 20, 2007

atqamar

I am unable to see your file, so what limit are you looking for?

$$\lim_{x \to 0^-} [2+x^{\frac{1}{3}}]^{\frac{1}{x}}$$ or

$$\lim_{x \to 0^+} [2+x^{\frac{1}{3}}]^{\frac{1}{x}}$$ or

$$\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^{\frac{1}{x}}$$ ?

Edit:
Just a thought- let $$p=\lim_{x \to \infty} \frac{1}{x}$$ and evaluate. Then substitute into $$\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^p$$ and evaluate.

Last edited: Oct 20, 2007
3. Oct 20, 2007

nrqed

As n goes to infinity, this goes to 1.

As n gets very large, you may approximate this by $n^{1/(3n)}$. taking the natural log of this (or log in any base), you can show that the ln goes to 0 as n goes to infinity. hence the initial function tends to e^0 = 1.

4. Oct 20, 2007

transgalactic

ooohh sorry i forgot a part in the limit

its

lim [2+n^(1/3)]^(1/n)
n>>infinity

and i did got

lim n^[1/(3n)]

you say i need to take log

it will be

log X =1/(3n)
n

i didnt understand how to solve it that way

Last edited: Oct 20, 2007
5. Oct 20, 2007

atqamar

Great.

Then, as I mentioned earlier, evaluate:
$$p=\lim_{x \to \infty} \frac{1}{x}$$

And plug into:
$$\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^p$$

Any non-zero base, raised to the power of $$0$$, is...

6. Oct 20, 2007

transgalactic

your first limit P gives me 0

the second one is infinity

and infinity^0 is not 1
its undefined

7. Oct 20, 2007

nrqed

You need to consider the limit of $\frac{\ln n}{n}$. The limit of this as n goes to infinity is zero. This is one of the basic limits and you probably can find the proof in yoru textbook.

8. Oct 20, 2007

ace123

For the second limit atqamar posted $$\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^p$$ is going to be 1 because anything other than zero raised to the 0 power is 1. How did you get undefined?

Edit: How did you get infinity to the zero power? Thats not what the base is going to be

Last edited: Oct 20, 2007
9. Oct 20, 2007

transgalactic

the base is :

2+ x^(1/3)

if x gose to infinity the the base gose to infinity

and p=0

it gives me (infinity)^0

and its undefined

10. Oct 20, 2007

dynamicsolo

It is undefined because this is what is called an "indeterminate power"; other examples are 0^0 and 1^infinity. Nonetheless there will be a value for it.

The solution involves one of the more exotic applications of L'Hopital's Rule, since this first has to be wrestled into the form of an indeterminate ratio 0/0 or infinity/infinity.

You start off by taking the logarithm of your expression, which will be

lim n-> inf. (1/n) ln [2 + n^(1/3)] ; this now has the apparent result 0 · infinity (an indeterminate product).

You next rewrite this product as a ratio, since f · g = f / (1/g) ; it is usually best to put the logarithmic expression in the numerator to apply L'Hopital's Rule, since we're going to be taking derivatives:

lim n-> inf. ( ln [2 + n^(1/3) ] ) / n .

This now has the apparent value infinity/infinity, making it a candidate for L'Hopital's Rule. Differentiate numerator and denominator separately:

lim n-> inf. [ ( 1/{ 2 + n^(1/3) } ) · ( 1/{ 3 · n^(2/3) } ) ] / 1 [using the Chain Rule on the numerator]

Since n appears in the denominator raised to positive powers, once this compound fraction is simplified, the limit is (1/2)·(1/3)·0 = 0 .

This, however, is the natural logarithm of the original expression, so this limit is the natural logarithm of our original limit. So the original limit is e^0 = 1 .

This is a formal technique for dealing with such problems; as nrqed and atqamar point out, though, this one can sort of be "eyeballed" for a solution...

Last edited: Oct 20, 2007