MHB What is the Limit as x Approaches 0 for the Absolute Value Function?

[karush]

$$\lim_{{x}\to{0}}\frac{\left| 2x-1 \right|-\left| 2x+1 \right|}{x}=-4$$

I tried /x but couldn't get the answer

 

[mathmari]

Since we have the absolute values we have to look at which points the expressions $2x-1$ and $2x+1$ change the sign.

We have that $2x-1>0$ for $x>\frac{1}{2}$ and $2x+1>0$ for $x>-\frac{1}{2}$.

Since we want to calculate the limit as $x \rightarrow 0$, we have to take the case $-\frac{1}{2}<x<\frac{1}{2}$.

So we have the following: $$\lim_{x \rightarrow 0}\frac{|2x-1|-|2x+1|}{x}=\lim_{x \rightarrow 0}\frac{-(2x-1)-(2x+1)}{x}\\ =\lim_{x \rightarrow 0}\frac{-2x+1-2x-1}{x}=\lim_{x \rightarrow 0}\frac{-4x}{x}\\ =\lim_{x \rightarrow 0}-4=-4$$

 

[karush]

I don't see why you choose the - sign

 

[mathmari]

We have the following identity of an absolute value:

$$|a|=\left\{\begin{matrix}
a \ \ ,& a>0\\
-a \ \ ,& a<0
\end{matrix}\right.$$

For $x>\frac{1}{2}$ we have that $2x-1$ is positive, so $|2x-1|=2x-1$.
For $x<\frac{1}{2}$ we have that $2x-1$ is negative, so $|2x-1|=-(2x-1)$.

Since we choose the interval $-\frac{1}{2}<x<\frac{1}{2}$ we have that $|2x-1|=-(2x-1)=-2x+1$.

 

[karush]

Got it

 

[Andrei1]

I suggest to prove the following:

$$\forall \epsilon>0\exists\delta>0\forall x\in\mathbb{R}^*\left(|x|<\delta\to\left|\frac{|2x-1|-|2x+1|}{x}+4\right|<\epsilon\right)$$

which defines

$$\lim_{x\to 0}\frac{|2x-1|-|2x+1|}{x}=-4.$$

 

[karush]

I suggest to prove the following:

$$\forall \epsilon>0\exists\delta>0\forall x\in\mathbb{R}^*\left(|x|<\delta\to\left|\frac{|2x-1|-|2x+1|}{x}+4\right|<\epsilon\right)$$

which defines

$$\lim_{x\to 0}\frac{|2x-1|-|2x+1|}{x}=-4.$$

Rather daunting!

 

[Greg]

L'Hopital's rule:

$$\lim_{x\to0}\dfrac{\left|2x-1\right|-\left|2x+1\right|}{x}=\lim_{x\to0}\dfrac{2(2x-1)}{\left|2x-1\right|}-\dfrac{2(2x+1)}{\left|2x+1\right|}=\lim_{x\to0}-2-2=-4$$

Using the chain rule, $$\dfrac{d}{dx}\,\left|f(x)\right|=f'(x)\dfrac{f(x)}{\left|f(x)\right|}$$.