Answer
(a) $\lim_{x\to1^+}\frac{\sqrt{2x}(x-1)}{|x-1|}=\sqrt2$
(b) $\lim_{x\to1^-}\frac{\sqrt{2x}(x-1)}{|x-1|}=-\sqrt2$
Work Step by Step
- For $x\gt1$: $$x-1\gt0\Rightarrow|x-1|=x-1$$
- For $x\lt1$: $$x-1\lt0\Rightarrow|x-1|=-(x-1)$$
a) $$A=\lim_{x\to1^+}\frac{\sqrt{2x}(x-1)}{|x-1|}$$
As $x\to1^+$, we only consider the values of $x$ to the right of $1$, which means $x\gt1$, so $|x-1|=x-1$.
Therefore, $$A=\lim_{x\to1^+}\frac{\sqrt{2x}(x-1)}{x-1}$$ $$A=\lim_{x\to1^+}\sqrt{2x}$$ $$A=\sqrt{2\times1}=\sqrt2$$
b) $$B=\lim_{x\to1^-}\frac{\sqrt{2x}(x-1)}{|x-1|}$$
As $x\to1^-$, we only consider the values of $x$ to the left of $1$, which means $x\lt1$, so $|x-1|=-(x-1)$.
Therefore, $$B=\lim_{x\to1^-}\frac{\sqrt{2x}(x-1)}{-(x-1)}$$ $$B=\lim_{x\to1^-}-\sqrt{2x}$$ $$B=-\sqrt{2\times1}=-\sqrt2$$
