- #1

alejandro7

- 13

- 0

limit when x->∞ (1+2^x)^(1/x)

I don't know how to proceed (I know I have to use l'Hospital's rule). It's a ∞^0 indetermination.

Thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter alejandro7
- Start date

In summary, the conversation discusses finding the limit of (1+2^x)^(1/x) as x approaches infinity, and suggests using L'Hopital's rule. The conversation also mentions using ln to simplify the expression and taking the exponent to find the final answer.

- #1

alejandro7

- 13

- 0

limit when x->∞ (1+2^x)^(1/x)

I don't know how to proceed (I know I have to use l'Hospital's rule). It's a ∞^0 indetermination.

Thanks!

Physics news on Phys.org

- #2

stephenkeiths

- 54

- 0

Try letting

[itex]y=ln((1+2^{x})^{\frac{1}{x}})[/itex]

Then what can you do??

[itex]y=ln((1+2^{x})^{\frac{1}{x}})[/itex]

Then what can you do??

Last edited:

- #3

alejandro7

- 13

- 0

1/x goes down.

- #4

stephenkeiths

- 54

- 0

now what form is it in? Can you use L'Hopitals rule now?

- #5

stephenkeiths

- 54

- 0

[itex]e^{y}=(1+2^{x})^{\frac{1}{x}}[/itex]

So when you find y, the limit of

[itex]y=ln((1+2^{x})^{\frac{1}{x}})[/itex]

you have to take [itex]e^{y}[/itex] to get the answer to the limit you're looking for.

- #6

alejandro7

- 13

- 0

Ok I have:

e^lim when x-> of ((ln(1-2^x)/x))

L'Hôpital now?

e^lim when x-> of ((ln(1-2^x)/x))

L'Hôpital now?

Last edited:

- #7

stephenkeiths

- 54

- 0

lim x-->∞ [itex]\frac{ln(1+2^{x})}{x}[/itex]

Then use L'Hopitals rule.

You will find the limit of this. To get the answer you want you have to exponentiate it (since you took the natural log in order to find it).

L'Hospital's Rule is a mathematical theorem used to solve indeterminate forms in calculus. It states that if the limit of a quotient of two functions is in an indeterminate form, then the limit can be evaluated by taking the derivative of the numerator and denominator separately and then finding the limit of the resulting quotient.

L'Hospital's Rule is used when the limit of a quotient of two functions is in an indeterminate form, such as 0/0 or ∞/∞. These indeterminate forms cannot be solved using basic algebraic methods, so L'Hospital's Rule is applied to evaluate the limit.

One common mistake when using L'Hospital's Rule is not checking if the indeterminate form is actually present. Sometimes, the limit can be evaluated using basic algebraic methods instead of using L'Hospital's Rule. It is also important to make sure that both the numerator and denominator of the original function are differentiable.

Yes, L'Hospital's Rule can be applied multiple times if the resulting limit is still in an indeterminate form. However, it is important to keep in mind that each time the rule is applied, the quotient becomes more complex and difficult to solve, so it is best to only use it as many times as necessary.

Yes, there are some limitations to using L'Hospital's Rule. It can only be applied to limits that are in indeterminate forms, and the functions involved must be differentiable. Additionally, it may not work if the functions involved are not continuous or if the limit is approaching a point of discontinuity.

- Replies
- 6

- Views
- 1K

- Replies
- 10

- Views
- 1K

- Replies
- 5

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 4

- Views
- 2K

- Replies
- 8

- Views
- 1K

- Replies
- 4

- Views
- 1K

- Replies
- 5

- Views
- 1K

- Replies
- 7

- Views
- 4K

- Replies
- 15

- Views
- 1K

Share: