$\begingroup$$$\lim_{x\to0}{\frac{|2x-1|-|2x+1|}{x}}=-4$$
Why? This came from a calculus book, before L'hopital is introduced. I couldn't find the answer myself, so I looked at the answers page. WolframAlpha agrees, and interestingly enough, the function is equal to $-4$ in the entire range $[-0.5,0.5]$, so maybe you could use squeeze theorem (which has been introduced) to evaluate the limit? Here is some of my working so far. $$\lim_{x\to0}{\frac{|2x-1|-|2x+1|}{x}}\\=2\lim_{x\to0}{\frac{|x-0.5|-|x+0.5|}{x}}\\=2\lim_{y\to0.5}{\frac{|y-1|-|y|}{y-0.5}}$$
The last step substitutes with $y=x+0.5$. It is the step at which I am stuck. Squeeze theorem? Thanks.
$\endgroup$23 Answers
$\begingroup$When $x$ is close enough to $0$, we can say that $-\frac12<x<\frac12$, which implies $2x-1<0$ and $2x+1>0$. Therefore, we can handle those absolute values, and rewrite the function as:
$$\frac{-(2x-1)-(2x+1)}{x} = \frac{-4x}{x}=-4$$
The limit of a constant function is simply that function's value.
$\endgroup$3$\begingroup$Hint: Try to see the limits in two cases where once $x \to 0^+$ and other $x \to 0^-$.
$\endgroup$$\begingroup$Hint:
As $x\to0,2x+1\to1,2x-1\to-1$
Now, for real $a,|a|=+a$ if $a\ge0$
and $=-a$ if $a<0$
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