Can calculus solve this?

You can do a Taylor expand around 0 of [itex]\sqrt{x+1}[/itex]

It is:
[tex]\sqrt{x+1}=1+x\left[\frac{1}{2\sqrt{x+1}}\right]_{x=0}+...[/tex]

 

The linear approximation of [itex]f(n)[/itex] about [itex]n_0[/itex] is given by:

[tex]L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)[/tex]


If [itex]f(n)=\sqrt{n+1}[/itex], then the linear approximation of f(n) around 0 is:

[tex]L(n)~=~f'(0)(n-0)+f(0)~ =~ \frac{1}{2\sqrt{0+1}}(n-0)+\sqrt{0+1}~ =~ \frac{n}{2}+1[/tex]

 

It's called linear approximation. Approximate your function with a line whose slope is the same as the original function's slope at x=0.
You find this slope from the derivative.

 

Even simpler, since

[tex] n + 1 \approx n + 1 + \frac {n^2} {4} = (n+\frac{1}{2})^2 [/tex]

if n<<1


[tex] \sqrt {n+1} \approx n + \frac{1}{2} [/tex]

 

You use a tangent line to approximate the function at [itex]n_0[/itex]. The equation of the tangent line is obviously given by [itex]y=f'(x_0)(x-x_0)+f(x_0)[/itex].

http://www.function-grapher.com/graphs/fc544b3a572de466f056ae4620193c3f.png [Broken]