Limit Question: Solving for (Infinity)^0 with n^(1/3) and (2+n^(1/3))^(1/n)

[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??

 

[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.

 

[nrqed]

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??

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.

 

[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

 

[atqamar]

ooohh sorry i forgot a part in the limit

its

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

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...

 

[transgalactic]

your first limit P gives me 0

the second one is infinity

and infinity^0 is not 1
its undefined

 

[nrqed]

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

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.

 

[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

 

[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

 

[dynamicsolo]

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

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...