Question

Prove the statement using epsilon and delta definition of a limit

lim as x -> -2 (x^2 -1) =3

Answer 1

are u typing or is it a mistake?

Answer 2

okay =)

Answer 3

Let epsilon > 0
Need to find a delta > 0 such that if |x - (-2)| < delta then | (x^2 - 1) - 3| < epsilon

|(x^2 - 1 )- 3| = |x^2 - 4| = |(x + 2)(x - 2)| < epsilon means
-epsilon < (x + 2)(x - 2)< epsilon
(-epsilon + 4)/(x - 2) < x + 2 < (epsilon + 4)/(x - 2)
-(epsilon + 4) < (-epsilon + 4)/(x - 2) < x + 2 < (epsilon + 4)/(x - 2) < epsilon + 4

==> |x + 2| < epsilon + 4
so if delta = epsilon + 4, then |x^2 - 4| < epsilon

Ans.

Since delta exists, then lim x^2 - 1 = 3 as x -> -2

Answer 4

can u plz help me w/ a limit problem? these are soo confusing!

Answer 5

its similar to this

Answer 6

ok

Answer 7

Trick is to focus on the antecedent (i.e. the second part of the if then part) and simplify until you can get the precedent (i.e. the first part of the if then statement)

Answer 8

correction in the "If p then q", p is the antecedent and q is the consequent. Focus on q to get to p.

Answer 9

anyway, what's the problem

Answer 10

We were learning about continuity. If f and g are continous functions with f(3) = 5 and
lim as x-> 3 [2 f(x) - g(x)] =4, find g(3).

Answer 11

lim 2f(x) - g(x)
= 2*lim f(x) - lim g(x)
= 2* f(3) - lim g(x)
= 2* 5 - lim g(x) = 4
2*5 - 4 = lim g(x) as x -> 3

lim g(x) = 10 -4 = 6 as x -> 3

Since lim g(x) as x->3 exists and g is continuous, this implies g(3) = lim g(x) as x->3 then g(3) = 6.

Answer 12

oh i see...tahnks alot for your help. it made it much more clearer. =)