Calculus 2A?? Limit Defintion Proof?

Question

use the definition of "lim (x > a) f(x) = L" to show that lim( x > 2) (x^2 + 1 = 5).

Answer 1

We will use d for delta and e for epsilon. The definition of
lim(x>2)(x^2 +1) = 5 is "for every e > 0, there exists d > 0 such that if 0 < abs(x - 2) < d and x is in the domain of f(x) = x^2 + 1, then abs (f(2) - 5) < e."

Let e > 0 be given. We must find d > 0. Since the domain of
f(x) = x^2 + 1 is the entire set of reals, there are no restrictions on x in 0 < abs(x - 2) < d except that x is not 2. We want

(1) abs(f(2) - 5)) = abs (x^2 - 4) = Abs(x - 2)*abs(x + 2) < e .

Notice that

(2) abs(x - 2) < d means -d < x - 2 < d, and thus

(3) 4 - d < x + 2 < d + 4.

From (1), (2), and (3), we want

Abs(x-2)*abs(x + 2)
Consider F(d) = d^2 + 4d - e. If this is negative, then we will have d^2 + 4d < e as required. But F(d) is a parabola which opens up (since the leading coefficient is > 0) , and this parabola will be negative when d is between its d-intercepts. It is easy to find the intercepts by solving the equation F(d) = 0. From the quadratic formula we have
d = (-4 +- sqrt(16 + 4e)/2 = -2 +- sqrt(4 + e) are the intercepts. Now we know that if
-2 - sqrt(4 + e) < d < -2 + sqrt(4 + e), then the parabole will be negative. In fact since d > 0, and the left intercept is negative, we choose 0 < d < -2 + sqrt(4 + e) .

In summary, when d > 0 is < -2 + sqrt(4 + e) and
0 < abs(x - 2) < d, then abs(f(2) - 5) < e.

Answer 2

Well I'm horrible at explaining these types of limit question but here goes.
Start with the definition of the limit

lim x → a f(x) = L
if and only if |f(x) - L| < ε whenever |x-a| < δ for ε,δ > 0

Apply this to the problem
lim x → 2 x^2+1 = 5
if and only if

|x^2 + 1 -5 | < ε whenever | x-2| < δ
|x^2 -4 | < ε whenever | x-2| < δ
|x-2||x+2| < ε whenever |x-2| < δ
Pretty straightforward so far, here's where it goes downhill.

If we can find some constant C such that
|x+2| < C then |x-2||x+2| < C|x-2| and we can make
c|x-2| < ε whenever |x-2 | < δ then
|x-2| < ε/c whenever |x-2 | < δ
then we can take δ = ε/c which proves the limit, as long as we can find such a C such that |x+2| < C

when we take limits we care about what is happening around a (in this case 2) so let's choose an interval 1 to the left and one to the right of 2

|x-2|<1 this makes x between 1 and 3
1 < x < 3
and therefore makes |x+3| between
4< |x+3| < 6 . Because of this limitation we can pick a C>|x+3| C=6 satisfies this (you could of picked anything above 6)
δ = ε/6

δ = min{1,ε/6} is how to express it meaning that you limited |x-2| to an interval of 1 and your δ = ε/6

Answer 3

My assumption, from the kinds of questions asked, is which you have this fact someplace in that section: "If lim f(x) and lim g(x) exist, then lim (f(x) + g(x) ) exists." we can then teach your fact by skill of contradiction. assume lim f(x) exists and that lim( f(x) + g(x) ) does no longer exist. additionally, assume that the suitable is pretend, so as that lim g(x) exists. Now, because of the fact lim f(x) exists and lim g(x) exists, we get that lim (f(x) + g(x)) exists, yet this contradicts our assumption. for this reason lim g(x) can no longer exist.