# Calculating limit using squeeze theorem

$$\lim\limits_{n\to\infty }\sqrt {n^2+(-1)^n}-n$$

First the domain , $$n \geq 1$$ since $$n^2+(-1)^n \geq n^2-1 \geq 0$$

let $$a_n =\sqrt {n^2+(-1)^n}-n$$ , so I wanted to simplify it by multiplying by the conjugate so $$\sqrt {n^2+(-1)^n}+n \geq 0$$ and I got

($$\sqrt {n^2+(-1)^n}-n) \cdot \frac{\sqrt {n^2+(-1)^n}+n}{\sqrt {n^2+(-1)^n}+n}$$ = $$\frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n}$$

here is where I was not sure of how to use the squeeze theorem and this is what I did

$$\frac{-1}{\sqrt{n^2-1}+n}\leq \frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n} \leq \frac{1}{\sqrt{n^2+1}+n}$$

let $$b_n = \frac{-1}{\sqrt{n^2-1}+n}$$ and $$c_n = \frac{1}{\sqrt{n^2+1}+n}$$

so $$\lim\limits_{n\to\infty }b_n = 0$$ and $$\lim\limits_{n\to\infty }c_n = 0$$ and according to squeeze theorem $$\lim\limits_{n\to\infty } a_n =0$$

but in the book it was solved this way :

after multiplying by the conjugate they also got $$\frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n}$$ then they did $$0 \leq|\frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n}| = \frac{|(-1)^n|}{|\sqrt{n^2+(-1)^n}+n|} = \frac{1}{\sqrt{n^2+(-1)^n}+n} \leq \frac{1}{n}$$ and then they evaluated the limit $$\lim\limits_{n\to\infty }0=0$$ and $$\lim\limits_{n\to\infty } \frac{1}{n} = 0$$ so according to squeeze theorem $$\lim\limits_{n\to\infty } a_n =0$$

What I did not understand in the book is why out of no where they used an absolute value? when am I allowed to do that?

and is my try also correct? or do I need to explain more stuff because the inequality is not necessarily correct because for example in $$C_n$$ the denominator actually increases so the value decreases (although I think that in this case it doesn't matter because the value is positive while $$b_n$$ is negative ) but I mean in general ?

Thank you

EDIT:

I tried solving it in another way , as a multiplication of a limit equal to zero and a bounded limit. If I can separate this limit into 2 limits $$\frac{(-1)^n}{\sqrt{n^2+(-1)^n}+n}$$ let $$b_n=(-1)^n$$ and $$c_n =\frac{1}{\sqrt{n^2+(-1)^n}+n}$$

$$b_n$$ is bounded of course $$|b_n|=1 \lt 2$$ and $$c_n =\frac{1}{\sqrt{n^2+(-1)^n}+n} \lt \frac{1}{n}$$ because as stated in the beginning the domain , $$n \geq 1$$ since $$n^2+(-1)^n \geq n^2-1 \geq 0$$ from here we get $$\sqrt{n^2-1} +n \geq n$$

therefore $$\lim\limits_{n\to\infty }b_n \cdot c_n=0$$

• There is a typo at the start when you multiply and divide by the conjugate. The square root sign shouldn't have the $-n$ under it. Nov 8, 2021 at 14:42

To address your question about how and why one can apply absolute values, here's a useful theorem:

A sequence $$(a_n)$$ converges to zero if and only if the corresponding sequence of absolute values $$(|a_n|)$$ also converges to zero.

So if your expectation is that $$\lim_{n \to \infty} a_n = 0$$, you can prove that by applying this theorem together with the squeeze theorem, in the following manner:

• find another sequence $$c_n$$ such that $$0 \le |a_n| \le c_n$$ and such that $$\lim_{n \to \infty} c_n = 0$$;
• apply the squeeze theorem to conclude that $$\lim_{n \to \infty} |a_n| = 0$$;
• apply the theorem above to conclude that $$\lim_{n \to \infty} a_n = 0$$

Here's an even more general form of the above theorem that is also useful:

A sequence $$(a_n)$$ converges to $$L$$ if and only if $$(a_n-L)$$ converges to zero, if and only if $$|a_n-L|$$ converges to zero.

• appreciate the detailed reply it is much more understandable . so if I expect the limit to be zero I can apply $0 \leq |a_n| \leq c_n$ and solve accordingly. so the thing I tried is not correct I suppose and I should try as you just explained? Nov 8, 2021 at 14:55
• Yes, but keep in mind the comment of @Digitallis. Nov 8, 2021 at 15:16
• It is worth mentioning that the first statement is really just the squeeze theorem in disguise. Namely, if $\vert a_n \vert \leq c_n$, then we can equivalently write $$-c_n \leq a_n \leq c_n$$ and then apply the squeeze theorem. Nov 8, 2021 at 15:27
• @LeeMosher Can you please check the small edit I made and tell me if this way is correct? Nov 8, 2021 at 16:54
• That looks fine. Nov 8, 2021 at 16:58