Back to MAT 194F

Term Test #1

Question 1

Use the epsilon, delta definition of a limit to show that
lim x->2 x**-2 = 0.25

Solution

Given any positive epsilon, we need to find a positive delta such that whenever x satisfies 0 < |x-2| < delta, it is true that |x**-2 - 0.25| < epsilon.

I claim that delta = min(1,(4/3)*epsilon) works.

If delta <= 1, then 1 < x < 3.

|x**-2 - 0.25|
= |x**-1 + 0.5| * |x**-1 - 0.5| [factoring]
< 1.5 * |x**-1 - 0.5| [x**-1<1]
= 1.5*|(2-x)/2x| [common denominator]
= 1.5*|(x-2)/2x| [negating within absolute value bars]
< 1.5*|(x-2)/2| [x>1]
= 0.75*|x-2|
< epsilon [delta<=(4/3)*epsilon]

Q.E.D.

Marking Scheme

Two points for general methodology, two points for correctly restricting delta to being less than or equal to a constant value, five points for correctly restricting delta to being less than or equal to a constant multiple of epsilon, one point for reaching the correct conclusion. Total: ten points.