For Chinu - (1)

As x $\rightarrow$ 0, |2x-1|= -(2x-1) and |2x+1|=2x+1..... $\small{\color{#624F41}{\text{(since |x|=x if x>0 and |x|=-x if x<0)}}}$ Hence limit simplifies to : $\Large \lim_{x\rightarrow 0} \frac{-2x+1-(2x+1)}{x}= \dfrac{-4x}{x}=\boxed{-4}$

If you consider the above graph when x approaches 0, the expression approaches -4.

Note that both the terms in the numerator are differentiable at zero. Thus we can apply L'Hôpital. By graphing these or otherwise, we see that the left term has derivative -2 at zero, and the right term has derivative +2 at zero. Hence the answer is -2-2=-4.

Let's rewrite the absolute value function as follows:

$\lim_{x\to 0}\dfrac{|2x-1|-|2x+1|}{x}=\lim_{x\to 0}\dfrac{\sqrt{(2x-1)^2}-\sqrt{(2x+1)^2}}{x}$

When evaluating at $x=0$ this comes out to be a $\frac{0}{0}$ indeterminate form, so we can use L'Hospital's rule:

$\lim_{x\to 0}\dfrac{\sqrt{(2x-1)^2}-\sqrt{(2x+1)^2}}{x}\stackrel{\text{H}}{=}\lim_{x\to 0}\dfrac{\frac{\text{d}}{\text{dx}}\sqrt{(2x-1)^2}-\sqrt{(2x+1)^2}}{\frac{\text{d}}{\text{dx}}x}$

$\Rightarrow\lim_{x\to 0}\dfrac{2(2x-1)}{\sqrt{(2x-1)^2}}-\dfrac{2(2x+1)}{\sqrt{(2x+1)^2}}=\boxed{-4}$

This question was in my calculus exam 1 month ago