How to evaluate the limit where x>>a?

Question

In physics, I was trying to evaluate an integral but in the end I had to evaluate a limit where x is far greater than a.

There was a solution, however, I can't seem to understand how they evaluate it.

Why are these expressions equal to each other?

Answer 1

They have only used binomial approximation, it is frequently used in physics.

If $\epsilon<<1$ then $(1+\epsilon)^r\approx 1+r\epsilon$. Here $\epsilon=\frac{a^2}{x^2}$

Answer 2

The function was approximated with it's first order taylor polinomial: $$(1+\varepsilon)^r\approx 1+r \varepsilon$$ When $\varepsilon$ is small enough. To be more precise, we could write $$(1+\varepsilon)^r=1+r\varepsilon+o(\varepsilon)$$ But the physicist tend to avoid the error term.

In your case, $\varepsilon=a^2/x^2$.

This approximation can also be derived from the definition of differentiability at $x=0$, i.e. it's the best linear approximation around $x=0$: $$f(\varepsilon)=f(0)+f'(0)\varepsilon+o(\varepsilon)$$

Answer 3

Hint: You have to show that $$\frac{1}{x}\left(1+\frac{a^2}{x^2}\right)^{-1/2}\approx \frac{1}{x}\left(1-\frac{a^2}{2x^2}\right)$$ for $$x>>a$$