# Solve Limit at Infinity: $(-1)^n \sqrt{n+1}/n$

• Potatochip911
In summary, the limit of the given expression as n approaches infinity is undefined. To solve limits at infinity, you can simplify the expression and use the rules of limits. L'Hopital's rule cannot be applied in this case. The (-1)^n term can affect the overall value of the expression. The expression can be rewritten to make it easier to evaluate the limit at infinity.
Potatochip911

## Homework Statement

$$\lim_{x\to\infty} \dfrac{(-1)^n\sqrt{n+1}}{n}$$

## Homework Equations

3. The Attempt at a Solution [/B]
This is what I managed to do but I just wanted to verify that this is the correct way of solving it, I'm mainly concerned about the fact that I took the absolute value with the log function, is that a valid operation?
$$y=\lim_{x\to\infty} \dfrac{(-1)^n\sqrt{n+1}}{n}$$
$$\ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}|$$
$$\ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}|$$
$$\ln y=\lim_{x\to\infty} \ln|(-1)^n|+\lim_{x\to\infty} \ln|\dfrac{\sqrt{n+1}}{n}|$$
$$\ln y=\lim_{x\to\infty} \dfrac{\ln|1|}{n^{-1}}+\lim_{x\to\infty} \ln|{\sqrt{1/n+1/n^2}}|$$
$$\ln y=\lim_{x\to\infty} \dfrac{0}{n^{-2}}+ \ln|0|$$
$$\ln y=-\infty$$
$$y=e^{-\infty}$$
$$y=0$$

Potatochip911 said:

## Homework Statement

$$\lim_{x\to\infty} \dfrac{(-1)^n\sqrt{n+1}}{n}$$

## Homework Equations

3. The Attempt at a Solution [/B]
This is what I managed to do but I just wanted to verify that this is the correct way of solving it, I'm mainly concerned about the fact that I took the absolute value with the log function, is that a valid operation?
$$y=\lim_{x\to\infty} \dfrac{(-1)^n\sqrt{n+1}}{n}$$
$$\ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}|$$
$$\ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}|$$
$$\ln y=\lim_{x\to\infty} \ln|(-1)^n|+\lim_{x\to\infty} \ln|\dfrac{\sqrt{n+1}}{n}|$$
$$\ln y=\lim_{x\to\infty} \dfrac{\ln|1|}{n^{-1}}+\lim_{x\to\infty} \ln|{\sqrt{1/n+1/n^2}}|$$
$$\ln y=\lim_{x\to\infty} \dfrac{0}{n^{-2}}+ \ln|0|$$
$$\ln y=-\infty$$
$$y=e^{-\infty}$$
$$y=0$$

There's a whole lot else wrong in there too. What's with the limit ##x \to \infty## when there is no ##x## in the expression? Assume you meant ##n \to \infty##. And there's not need to take a log to begin with. You seem ok with ##lim_{n \to\infty} \dfrac{\sqrt{n+1}}{n}=0##. The ##(-1)^n## doesn't change that much. Just use a squeeze argument.

Dick said:
There's a whole lot else wrong in there too. What's with the limit ##x \to \infty## when there is no ##x## in the expression? Assume you meant ##n \to \infty##. And there's not need to take a log to begin with. You seem ok with ##lim_{n \to\infty} \dfrac{\sqrt{n+1}}{n}=0##. The ##(-1)^n## doesn't change that much. Just use a squeeze argument.
Could you explain what's wrong other than the fact I accidentally used x->infinity so I don't make that mistake again?

Potatochip911 said:
Could you explain what's wrong other than the fact I accidentally used x->infinity so I don't make that mistake again?

Not as much as I thought at first, but the limit of ##x_n## is not necessarily the same as ##|x_n|##. I would just skip the log and absolute value in the argument altogether.

Ignoring the (-1)n factor for the moment, you have
##\frac{\sqrt{n+1}}{n} = \frac{\sqrt{n}\sqrt{1 + 1/n}}{\sqrt{n}\sqrt{n}} = \frac {\sqrt{1 + 1/n}}{\sqrt{n}}##
Can you take the limit now?

For the original problem, use the squeeze theorem that Dick suggests. I agree that logs and absolute values are not needed.

## 1. What is the limit of the given expression as n approaches infinity?

The limit of the given expression as n approaches infinity is undefined.

## 2. How do you solve limits at infinity?

To solve limits at infinity, you need to simplify the expression by factoring, rationalizing, or using algebraic manipulation. Then, you can use the rules of limits, such as the limit of a quotient, to evaluate the limit as n approaches infinity.

## 3. Can you use L'Hopital's rule to solve this limit at infinity?

No, L'Hopital's rule can only be used for limits involving indeterminate forms such as 0/0 or ∞/∞. In this case, the limit does not involve an indeterminate form and therefore L'Hopital's rule cannot be applied.

## 4. What is the significance of (-1)^n in the given expression?

The (-1)^n term alternates between positive and negative values as n increases, which can affect the overall value of the expression. It is important to consider this term when evaluating the limit at infinity.

## 5. Is there a way to rewrite the expression to make it easier to solve the limit at infinity?

Yes, the expression can be rewritten as (-1)^n (n+1)^1/2/n, which can then be simplified using the rules of limits. This can make it easier to evaluate the limit at infinity.

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