Finding limit of sequence, done right: $a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$?

Question

I need to find the limit of this sequence:

$a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$

So I multiply with this since $(a-b)(a+b)=(a^2-b^2)$

$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$

And get $\dfrac{9-n+9}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$

Then divide by n?

And get $\dfrac{9/n-n/n+9/n}{\sqrt{n^2/n+9/n}+\sqrt{n^2/n-n+9/n}}$

$\lim = \dfrac{1}{\infty} = 0$?

Answer 1

$$\dfrac{n}{(\sqrt{n^2+9}+\sqrt{n^2-n+9})}=\dfrac{n/n}{\frac{1}{n}(\sqrt{n^2+9}+\sqrt{n^2-n+9})}$$ $$=\dfrac{1}{\sqrt{n^2/n^2+9/n^2}+\sqrt{n^2/n^2-n/n^2+9/n^2}}$$

Answer 2

After multiplying the initial equation by $$\dfrac{\sqrt{n^2+9}+\sqrt{n^2-n+9}}{\sqrt{n^2+9}+\sqrt{n^2-n+9}}$$ You should get $(n^2+9) - (n^2 - n +9) = n$ in the numerator, i.e., you should have this:

$$\frac{n}{\sqrt{n^2 + 9}+\sqrt{n^2-n+3}}$$

Now try dividing numerator and denominator by $n$:

$$\lim_{n\to \infty} \frac n{\sqrt{n^2 + 9}+\sqrt{n^2-n+3}}= \lim_{n\to \infty} \frac 1{\sqrt{1+\frac 9{n^2}} +\sqrt{1-\frac 1n+\frac 9{n^2}}} = \frac 12$$