# Limits

• Apr 24th 2010, 01:42 AM
SyNtHeSiS
Limits
Deduce that if |x -3| < 1, then |(x + 2)^(2) - 25| < 11|x - 3|

Use this result to show, without using the Limit Laws, that

lim x -> 3 of (x + 2)^(2) = 25

I am really confused and dont understand this question
• Apr 24th 2010, 01:52 AM
simplependulum
Quote:

Originally Posted by SyNtHeSiS
Deduce that if |x -3| < 1, then |(x + 2)^(2) - 25| < 11|x - 3|

Use this result to show, without using the Limit Laws, that

lim x -> 3 of (x + 2)^(2) = 25

I am really confused and dont understand this question

\$\displaystyle |x-3|<1 \$

\$\displaystyle |(x+2)^2 - 25| = |(x+2)^2 - 5^2| = |(x+2-5)(x+2+5)|\$

\$\displaystyle = |(x-3)(x+7)| \$

Also from \$\displaystyle |x-3|<1 \$ , we have \$\displaystyle 2<x<4 \$ so \$\displaystyle 9 < x+7 < 11 \$

Therefore , \$\displaystyle |(x+2)^2 - 25| = |x-3||x+7| < 11|x-3|\$

For the second part , we have to prove it by checking the definition ... What similarity between part 1 and the definition can you find ?
• May 7th 2010, 04:06 PM
SyNtHeSiS
How are you suppose to know that it is suppose to be manipulated in that way? I would never have thought of that (Surprised)