# How do you evaluate a limit n->inf. if there's a (-1)^n term?

• pylauzier
In summary, the person is asking for help with evaluating the limit lim n->∞ (-1)n *sqrt(n+1)/n, and someone suggests using the alternating series test. The person then asks about changing the limit to lim n->infinity (-1)^n*sqrt(n^2+1)/n. It is determined that there is no limit for this expression.
pylauzier

## Homework Statement

I decided to work through my old calculus notes and I can't find any information about these limits. Basically I'm trying to evaluate lim n->∞ (-1)n *sqrt(n+1)/n.

## Homework Equations

Not really applicable.

## The Attempt at a Solution

Well, in this particular case it's easy to see that the limit will be zero if we divide the numerator and denominator by n, since the square root will tend towards zero. However, could this limit have been evaluated if the squared n was actually n2?

You should be able to handle that one with the alternating series test. Now how exactly are you thinking about changing it?

What do you mean by "Now how exactly are you thinking about changing it?" ??

I mean this part. "However, could this limit have been evaluated if the squared n was actually n^2?". What do you mean by that?

Oh, basically I said that I managed to solve the first limit I posted since the numerator tends toward 0 as n -> ∞, so that one was obvious. However with n2 instead of n under the square root, the limit would be +/- 1. If I recall correctly that's not a valid answer, so surely there's a way to evaluate further?

pylauzier said:
Oh, basically I said that I managed to solve the first limit I posted since the numerator tends toward 0 as n -> ∞, so that one was obvious. However with n2 instead of n under the square root, the limit would be +/- 1. If I recall correctly that's not a valid answer, so surely there's a way to evaluate further?

If you are thinking about lim n->infinity (-1)^n*sqrt(n^2+1)/n then there is no limit. It alternates between points near +1 and -1, as you said.

Dick said:
If you are thinking about lim n->infinity (-1)^n*sqrt(n^2+1)/n then there is no limit. It alternates between points near +1 and -1, as you said.

Alright that answers my question, thanks!

## 1. What is a limit?

A limit is a concept in calculus that describes the behavior of a function as the input or independent variable approaches a certain value. It is used to determine the overall trend or behavior of a function, even if the function is not defined at that particular value.

## 2. How do you evaluate a limit as n approaches infinity?

To evaluate a limit as n approaches infinity, you can take the limit of the function as n gets larger and larger. This means plugging in larger and larger values for n and observing the resulting values of the function. If the function approaches a specific value as n gets larger, then that value is the limit.

## 3. What does the (-1)^n term mean in a limit?

The (-1)^n term is a sequence that alternates between positive and negative values as n increases. It is commonly used in limits to represent an alternating pattern in the function.

## 4. How does the (-1)^n term affect the evaluation of a limit?

The (-1)^n term can affect the evaluation of a limit by causing the function to oscillate between positive and negative values as n gets larger. This can make it difficult to determine the overall trend or behavior of the function, and may require additional techniques, such as the squeeze theorem, to evaluate the limit.

## 5. Are there any special cases when evaluating a limit with a (-1)^n term?

Yes, there are some special cases when evaluating a limit with a (-1)^n term. For example, if the function is a polynomial with an odd degree, the limit as n approaches infinity will depend on the sign of the leading coefficient. If the leading coefficient is positive, the limit will be positive infinity, and if the leading coefficient is negative, the limit will be negative infinity. Additionally, if the function has a repeating decimal pattern, the limit may not exist as n approaches infinity.

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