Why is $\sqrt{1 + x^2}$ approximately equal to $1 + \frac{x^2}{2}$?

Question

I saw this in Shankar’s Physics book and couldn’t make out the reasoning behind it. I would assume the dx and derivative have nothing to do with it.

https://i.stack.imgur.com/D21AW.jpg

Answer 1

Notice that if $f(x) = \sqrt{1+x}$ , then $f'(x) = \dfrac{1}{2 \sqrt{1+x}}$. Moreover, the equation of the tangent line of $f(x)$ at $x=0$ is given by

$$ y- f(0) = f'(0)(x-0) \implies y = 1 + \frac{1}{2}x $$

Therefore, for values of $x \approx 0$, we have

$$ \sqrt{1+x} \approx 1 +\frac{1}{2} x $$

in particular, if we replace $x$ with $x^2$, one has

$$ \boxed{ \sqrt{1+x^2} \approx 1 +\frac{1}{2} x^2 }$$

Answer 2

$\sqrt{1+x} \approx 1+x/2$ is the linear approximation of $\sqrt{1+x}$ near $x=0$, which is based on the fact that $\left. \frac{d}{dx} \sqrt{1+x} \right |_{x=0}=1/2$.

Answer 3

Source: How to find square roots

Here $x$ is close to $0$.

Imagine you have a square of area $1+x^2$. The first estimate of the side length is about $1$, sounds reasonable?

But then that's an underestimation of that side length, because the estimated $1\times1$ square $\square$ is missing an L-shaped region $\blacksquare$, whose area is $x^2$!

$$\left.\begin{array}{ccccccc} \square&\square&\square&\square&\square&\square&\blacksquare\\ \square&\square&\square&\square&\square&\square&\blacksquare\\ \square&\square&\square&\square&\square&\square&\blacksquare\\ \square&\square&\square&\square&\square&\square&\blacksquare\\ \square&\square&\square&\square&\square&\square&\blacksquare\\ \square&\square&\square&\square&\square&\square&\blacksquare\\ \blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare&\blacksquare\\ \end{array}\quad\right\}\sqrt{1+x^2}=1+\ldots?$$

Then the next estimate is to pretend that the L-shaped region is just two $1\times \frac{x^2}2$ rectangles. This give an over-estimated side length of the square

$$l = \sqrt{1+x^2} \approx 1+\frac{x^2}2$$