Finding limit of sequence, done right: $a(n)=\sqrt{n^2+9}-\sqrt{n^2-n+9}$? 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 $\dfrac{9/n-n/n+9/n}{\sqrt{n^2/n+9/n}+\sqrt{n^2/n-n+9/n}}$
$\lim = \dfrac{1}{\infty} = 0$?

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