Show that function approximates another function for large values of R I am not sure how to approach this problem that came up during the semester that I never did get correct (studying for finals now). It is calculus class and I can't figure out which calculus technique I should be applying here. A quick direction on how to approach this would be appreciated!
Given:
$$V = 2\pi\sigma\left(\sqrt{R^2+a^2}-R\right)$$
where $a$ and $\sigma$ are constants. Show that for large $R$ that:
$$V \approx \frac{\pi a^2\sigma}{R}$$
Hint:
Write $\left(\sqrt{R^2+a^2}-R\right) = R\left(\sqrt{1+a^2R^{-2}}-1\right)$ and think of $a^2R^{-2}$ as small.
 A: Do you know about Taylor series? The Taylor expansion of $\sqrt{1+x}$ if $|x|<1$ is:
$$\sqrt{1+x}=1+\dfrac{1}{2}x+xf(x)$$
where $f(x)\rightarrow 0$ as $x\rightarrow 0$. In our case, assume $a^2R^{-2}<1$ that is $R$ is large enough and we get:
$$\sqrt{R^2+a^2}-R=R\big(\sqrt{1+a^2R^{-2}}-1\big)=R\big(1+\dfrac{1}{2}a^2R^{-2}+a^2R^{-2}f(a^2R^{-2})-1\big)$$
so that
$$\sqrt{R^2+a^2}-R=\dfrac{a^2}{2R}+\dfrac{a^2}{R}f(a^2R^{-2})$$
Properties of $f$ ensure that
$$\dfrac{a^2}{2R}f(a^2R^{-2})\longrightarrow 0$$
as $R\rightarrow\infty$ so that 
$$\sqrt{R^2+a^2}-R\simeq \dfrac{a^2}{2R}$$
Your formula follows.
A: $$2\pi\sigma\left(\sqrt{R^2+a^2}-R^2\right)\cdot\frac{\sqrt{R^2+a^2}+R^2}{\sqrt{R^2+a^2}+R^2}=\frac{2\pi\sigma a^2}{\sqrt{R^2+a^2}+R^2}=$$
$$=\frac{2\pi\sigma a^2}R\cdot\frac1{\sqrt{1+\frac{a^2}{R^2}}+1}\cong\frac{\pi \sigma a^2}R$$
Because
$$\frac1{\sqrt{1+\frac{a^2}{R^2}}+1}\xrightarrow[R\to\infty]{}\frac1{1+1}=\frac12$$
A: Let us use the following approximation
for small $x$:
$$(1+x)^\alpha\approx 1+\alpha x$$
then for large $R$,
$$\sqrt{R^2+a^2}=R(1+\frac{a^2}{R^2})^\frac{1}{2}\approx R(1+\frac{a^2}{2R^2})$$
thus
$$\sqrt{R^2+a^2}-R\approx \frac{a^2}{2R}$$.
and
$$V\approx \frac{\pi a^2\sigma}{R}$$
