can someone help me with a calculus problem?

Question

its the Limit as X approaches -2 for the absolute value of X+2 when it equals 0.

kinda looks like this:

Lim [ X+2 ] =0
x=>-2

it needs to be done using the precise definition of a limit

Answer 1

Ah, ε-δ stuff. Okay, we wish to show that [x→-2]lim |x+2| = 0. This means that we need to prove that ∀ε>0, we can find δ>0 such that 0 < |x-(-2)| < δ → | |x+2| - 0| < ε. Of course, this is extremely easy, since it suffices in all cases to let δ=ε. Then 0 < |x-(-2)| < δ → |x+2| < δ → | |x+2| | < δ → | |x+2| - 0| < δ → | |x+2| - 0| < ε. The only thing you actually need to know here is that the absolute value operator is idempotent -- that is, applying it twice yields the same result as applying it once.

Answer 2

For this problem, the function x+2 is continuous everywhere, so to find the limit, simply plug in -2 to X+2 to receive your answer:

0

Answer 3

I am not sure you wrote this question out correctly. limit equations rarely have answers. However, if f(x) = x+2, the limit of f(x) as x -> 2 is 0.

Answer 4

If f(x) = x + 2 then the formula to differentiate from 1st principle is given by:-
f `(x) = lim h->0 (f(x + h) - f (x)) / h
f ` (x) = lim h ->0 ((x + 2 + h - (x + 2)) / h
f ` (x) = lim h->0 (h / h)
f ` (x) = 1
(which is probably completely irrelevant to what you are asking!?)

Answer 5

You can solve this by directly substituting the value the limit is appraching for x

The limit is 0, which is already given in your question, so I guess I'm not sure what you're asking.

Answer 6

Lim |x+2| =|-2+2| = 0
x->-2