Algebra of limits - proving limit is 2 I'm working on a question:

Not sure how to begin - what should I multiply by first? If I multiplied by $\dfrac{\sqrt{n+1}-\sqrt{n-1}}{\sqrt{n+1}-\sqrt{n-1}}$ would the difference of two squares give $(n+1)-(n-1)$ in the numerator and so, 2? If this is the correct first step to make, then what do I do next?
 A: Hint: write as
$$\frac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt{n}}=\frac{\sqrt{n}(\sqrt{1+1/n}+\sqrt{1-1/n})}{\sqrt{n}}=\sqrt{1+1/n}+\sqrt{1-1/n}$$
A: Divide the top and bottom by $\sqrt{n}$ and it should become clear.
A: I would work separately
with the two terms
and show that
$\lim \frac{\sqrt{n+1}}{\sqrt{n}}
=1
$
and
$\lim \frac{\sqrt{n-1}}{\sqrt{n}}
=1
$.
For the first,
$\frac{\sqrt{n+1}}{\sqrt{n}}-1
=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}
=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n}(\sqrt{n+1}+\sqrt{n})}
=\frac{(n+1)-n}{\sqrt{n(n+1)}+n}
=\frac{1}{\sqrt{n(n+1)}+n}
$
and this obviously goes to zero.
For the second,
just replace the "$+1$" with "$-1$".
More generally,
you can show that,
for any real $a$,
$\lim_{n \to \infty}\frac{\sqrt{n+a}}{\sqrt{n}}
=1
$
in a similar way:
If $n > |a|$,
$\frac{\sqrt{n+a}}{\sqrt{n}}-1
=\frac{\sqrt{n+a}-\sqrt{n}}{\sqrt{n}}
=\frac{(\sqrt{n+a}-\sqrt{n})(\sqrt{n+a}+\sqrt{n})}{\sqrt{n}(\sqrt{n+a}+\sqrt{n})}
=\frac{(n+a)-n}{\sqrt{n(n+a)}+n}
=\frac{a}{\sqrt{n(n+1)}+n}
$
and this obviously goes to zero
as $n \to \infty$.
Even more generally,
you can show that,
for any real function
$a(n)$
such that
$\lim_{n \to \infty} \frac{a(n)}{n}
=0$,
$\lim_{n \to \infty}\frac{\sqrt{n+a(n)}}{\sqrt{n}}
=1
$
in the same way:
If $n > |a(n)|$ for $n > N$,
$\frac{\sqrt{n+a(n)}}{\sqrt{n}}-1
=\frac{\sqrt{n+a(n)}-\sqrt{n}}{\sqrt{n}}
=\frac{(\sqrt{n+a(n)}-\sqrt{n})(\sqrt{n+a(n)}+\sqrt{n})}{\sqrt{n}(\sqrt{n+a(n)}+\sqrt{n})}
=\frac{(n+a(n))-n}{\sqrt{n(n+a(n))}+n}
=\frac{a(n)}{\sqrt{n(n+a(n))}+n}
$
and this obviously goes to zero
as $n \to \infty$.
