Finding limit (or rather simplifying) a specific function

Question

The question is asking to find the limit of: $$\lim_{x \to 0}\frac{\sqrt{1+x} - \sqrt{1-x}}{x}$$If we plug x into the equation, both the denominator and numerator become 0 so I have to simplify the equation somehow. The answer according to the textbook is 1 although I have no clue as to how I would get that answer myself. Any hints are greatly appreciated!

Answer 1

Hint: $$\frac{\sqrt{1+x}-\sqrt{1-x}}{x}=\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})}=\frac{1+x-(1-x)}{x(\sqrt{1+x}+\sqrt{1-x})}=\frac{2}{\sqrt{1+x}+\sqrt{1-x}}.$$

Answer 2

Let $f_\alpha(x) = \sqrt{1+\alpha x}$. Then $f_\alpha (0) = 1$, $f_\alpha' (0) = \frac{\alpha}{2}$.

Now note that $\frac{\sqrt{1+x} - \sqrt{1-x}}{x} = \frac{(f_1(x)-f_1(0)) - ( f_{-1}(x)-f_{-1}(0)) }{x} $, hence $\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x} = \lim_{x \to 0} ( \frac{f_1(x)-f_1(0) }{x} - \frac{ f_{-1}(x)-f_{-1}(0) }{x} ) = \frac{1}{2} - \frac{-1}{2} = 1$.

Answer 3

If you are allowed to use approximations coming from Taylor expansions,

Sqrt[1+x] is approximated by (1 + x / 2 + ....) ; Sqrt[1-x] is approximated by (1 - x / 2 + ....)

So, the numerator is (1 + x /2) - (1 - x /2) = x and then the result of 1

Answer 4

$\displaystyle{\large\sqrt{1 \pm x\,}\, \approx 1 \pm {1 \over 2}\,x\quad}$ when $\displaystyle{\large\quad\left\vert x\right\vert \approx 0}$.

$$ {\sqrt{\, 1 + x\,} - \sqrt{\,1 - x\,} \over x} \quad\to\quad {\left(1 + x/2\right) - \left(1 - x/2\right) \over x} = \color{#ff0000}{\large 1} \quad\mbox{when}\quad \color{#ff0000}{\large x \to 0} $$

$\color{#ff00ff}{\large\bf\mbox{Something like that could lead easily to a fallacy !!!}\,.}$ For example \begin{align} \lim_{x \to 0}{\sqrt{\, 1 + x\,} - \sqrt{\,1 - x\,} \over x} &= \lim_{x \to 0} {\left(\overbrace{\ \sqrt{\, 1 + x\,} - 1\ }^{\to\ 0}\ +\ \mu x/2\right) - \left(\overbrace{\ \sqrt{\,1 - x\,} - 1}^{\to\ 0} - \mu x/2\right)\over x} \\[3mm]&= \lim_{x \to 0}{\left(\mu x/2\right) - \left(-\mu x/2\right)\over x} = \mu \end{align} And then, the limit can take any value we choose !!!.