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

[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?
$$y=\lim_{x\to\infty} \dfrac{(-1)^n\sqrt{n+1}}{n} $$
$$ \ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}| $$
$$ \ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}| $$
$$\ln y=\lim_{x\to\infty} \ln|(-1)^n|+\lim_{x\to\infty} \ln|\dfrac{\sqrt{n+1}}{n}|$$
$$\ln y=\lim_{x\to\infty} \dfrac{\ln|1|}{n^{-1}}+\lim_{x\to\infty} \ln|{\sqrt{1/n+1/n^2}}|$$
$$\ln y=\lim_{x\to\infty} \dfrac{0}{n^{-2}}+ \ln|0|$$
$$\ln y=-\infty$$
$$y=e^{-\infty}$$
$$y=0$$

 

[Dick]

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.

 

[Potatochip911]

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?

 

[Dick]

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.

 

[Mark44]

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.