Proof: $(a^2+h)^{1/2} \approx a+\frac{h}{2a}$ for $0<h<a^2$

Question

Proof with binomial theorem:

$(a^2+h)^{1/2} \approx a+\frac{h}{2a}$ for $0<h<a^2$

Solution:

$(a^2+h)^{1/2} = a(1+\frac{h}{a^2})^{1/2}$

I think there must be an additional step between the last and second last step? But which one?

How do I proof that it's only value for for $0<h<a^2$?

Answer 1

We ignore higher order terms, i.e. terms where power of $h$ is at least $2$.

The approximation is better the smaller $\frac{h}{a^2}$ is.

Answer 2

Square the RHS. You'll get the LHS + higher-order. In that region you can then argue that the higher-order terms are "small"

Answer 3

You could then try Bernoulli's inequality $$(a^2+h)^{1/2}=a(1+h/a^2)^{1/2}\ge a\left(1+\frac {h}{2a^2}\right)$$

So could state as $$(a^2+h)^{1/2}\approx a+\frac {h}{2a}$$

By the way by using calculus for this question,the method stated by you is known as Linearisation and is done because the higher powers of binomial terms make the terms become negligible.