Limits Plus: I

Calculus Level 3

Evaluate without the use of L'Hôpital's Rule:

$\lim _{ x\rightarrow 0 }{ \frac { \left| 2x-1\right| -\left| 2x+1\right| }{x } } .$

Problem credit: Calculus: 6E, James Stewart

1 solution

John M.
Aug 29, 2014

For $-\frac{1}{2}<x<\frac{1}{2}$ , we have $2x-1<0$ and $2x+1>0$ , so $\left| 2x-1 \right| =-(2x-1)$ and $\left|2x+1 \right| =2x+1$ .

Therefore,

$\lim _{ x\rightarrow 0 }{ \frac { \left| 2x-1\right| -\left| 2x+1\right| }{x } }$

$=\lim _{ x\rightarrow 0 }{ \frac {(-2x-1)-(2x+1) }{x } }$

$=\lim_{x\rightarrow 0}{\frac{-4x}{x}}$

$=\lim_{x\rightarrow 0}{-4}$

$=\boxed{-4}$ .

Basic use of the property of modulus.

Anurag Pandey - 6 years, 1 month ago