- #1
quanticism
- 29
- 3
Homework Statement
Show that
[tex]
\lim_{x\to2}\frac{x+1}{x+2}=\frac{3}{4}.
[/tex]
Homework Equations
Let
[tex]\epsilon>0 [/tex]. We seek a number [tex]\delta>0[/tex]: if [tex]|x-2|<\delta[/tex] then [tex]|\frac{x+1}{x+2}-\frac{3}{4}|<\epsilon.[/tex]
The Attempt at a Solution
Now
[tex]
|\frac{x+1}{x+2}-\frac{3}{4}|=|\frac{4x+4-3x-6}{4(x+2)}|=|\frac{x-2}{4(x+2)}|.
[/tex]
So for
[tex]
|\frac{x+1}{x+2}-\frac{3}{4}|<\epsilon[/tex], we require [tex]|\frac{x-2}{4(x+2)}|<\epsilon.
[/tex]
ie.
[tex]
|\frac{x-2}{x+2}|<4\epsilon.
[/tex]
I got stuck here since I'm not sure how to express
[tex]
|\frac{x-2}{x+2}|
[/tex]
in terms of |x-2|