TITLE: SQUEEZE THEOREM, LEFT HAND SIDE AND RIGHT AHND SIDE LIMITS
PROBLEM 1: Squeeze Theorem
Given 2๐ฅโ2โค๐(๐ฅ)โค๐ฅ2โ2๐ฅ+2, ๐ฅโฅ 0.
FIND: lim
๐ฅโ2๐(๐ฅ)
SOLUTION: This is a perfect problem where we can use the Squeeze Theorem which states that if we
have ๐(๐ฅ) โค ๐(๐ฅ) โค โ(๐ฅ) and lim
๐ฅโ๐๐(๐ฅ)=๐ฟ= lim
๐ฅโ๐โ(๐ฅ) then the lim
๐ฅโ๐๐(๐ฅ)=๐ฟ. In our problem, we just
need to solve for the lim
๐ฅโ2(2๐ฅโ2) and lim
๐ฅโ2(๐ฅ2โ2๐ฅ+2) and show that they are equal. Thus,
i. lim
๐ฅโ2(2๐ฅโ2)=2(2)โ2 - Apply limit by replacing x=2
= 4โ2 - Simplify
= ๐
ii.
lim
๐ฅโ2(๐ฅ2โ2๐ฅ+2)= (2)2โ2(2)+2
=4โ4+ 2
= 2
Since the two limits are equal,
lim
๐ฅโ2(2๐ฅโ2)โคlim
๐ฅโ2๐(๐ฅ)โคlim
๐ฅโ2(๐ฅ2โ2๐ฅ+2)
2 โค lim
๐ฅโ2๐(๐ฅ)โค 2.
By Squeeze Theorem, ๐ฅ๐ข๐ฆ
๐โ๐๐(๐)=๐.
PROBLEM 2. Right Hand Side and Left Hand Side Limit
GIVEN: ๐(๐ฅ)=๐ฅ2+๐ฅโ2
|๐ฅโ1|
FIND: lim
๐ฅโ1๐(๐ฅ)
SOLUTION:
In this case, notice that there is an absolute value involved, so we need to break this down into
two as follows:
๐(๐ฅ)=
{
๐ฅ2+๐ฅโ2
โ(๐ฅโ1) ๐๐ ๐ฅ<1
๐ฅ2+๐ฅโ2
(๐ฅโ1) ๐๐ ๐ฅ โฅ 1
