Analyze the existence of the following limit:
[tex]\lim _{x\to2}x^2-\frac{2|x-2|}{x-2}[/tex]To better analyze the limit, we'll use some properties to isolate the required part from the others:
[tex]\lim _{x\to2}x^2-\frac{2|x-2|}{x-2}=\lim _{x\to2}x^2-\lim _{x\to2}\frac{2|x-2|}{x-2}[/tex]The first limit is easy to calculate, it's just to replace the value of x. The second limit will be further worked on:
[tex]2^2-2\lim _{x\to2}\frac{|x-2|}{x-2}=4-2\lim _{x\to2}\frac{|x-2|}{x-2}[/tex]Now focus on this part:
[tex]\lim _{x\to2}\frac{|x-2|}{x-2}[/tex]We'll approach the value of x=2 by using the set
X={-0.03, -0.02, -0.01, 0, 0.01, 0.02, 0.03}
Note these values are the infinitesimal approaches to the required value of x=2, thus the values of x to use are x={1.97, 1.98, 1.99, 2, 2.01, 2.02, 2.03}
[tex]\frac{|1.97-2|}{1.97-2}=\frac{|-0.03|}{-0.03}=\frac{0.03}{-0.03}=-1[/tex][tex]\frac{|1.98-2|}{1.98-2}=\frac{|-0.02|}{-0.02}=\frac{0.02}{-0.02}=-1[/tex][tex]\frac{|1.99-2|}{1.99-2}=\frac{|-0.01|}{-0.01}=\frac{0.01}{-0.01}=-1[/tex]The value of x=2 cannot be used because it would produce the division 0/0 and it's undefined.
[tex]\frac{|2.01-2|}{2.01-2}=\frac{|0.01|}{0.01}=\frac{0.01}{0.01}=1[/tex][tex]\frac{|2.02-2|}{2.02-2}=\frac{|0.02|}{0.02}=\frac{0.02}{0.02}=1[/tex][tex]\frac{|2.03-2|}{2.03-2}=\frac{|0.03|}{0.03}=\frac{0.03}{0.03}=1[/tex]We can see this limit results in -1 for the negative values of X and 1 for the positive values of X. Since the limits are not equal, the limit does not exist
