Question

Iim √1+х - √1-х x-o 3х

Answer 1

$\displaystyle \lim_{x \to 0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{3x}=\left\{\dfrac{0}{0}\right\}=\lim_{x \to 0}\dfrac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{3x(\sqrt{1+x}+\sqrt{1-x})}=\\ \\ \\ =\lim_{x \to 0}\dfrac{\left(\sqrt{1+x}\right)^2-\left(\sqrt{1-x}\right)^2}{3x\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\lim_{x \to 0}\dfrac{1+x-1+x}{3x\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\\ \\ \\ =\lim_{x \to 0}\dfrac{2x}{3x\left(\sqrt{1+0}+\sqrt{1-0}\right)}=\dfrac{2}{3\cdot 2}=\dfrac{1}{3}$

Answer 2

To simplify the expression, let's break it down step by step:

Given expression: √(1+x) - √(1-x) × (x - 3x)

Step 1: Simplify the expression inside the brackets.

√(1+x) - √(1-x) × (x - 3x) = √(1+x) - √(1-x) × (-2x) [Simplified x - 3x to -2x]

Step 2: Distribute the multiplication.

√(1+x) - √(1-x) × (-2x) = √(1+x) + 2x√(1-x) [Distributed the negative sign]

So, the simplified expression is √(1+x) + 2x√(1-x).