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Calculus/Calculus (LIMITS)

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Question
I want the proof of lim x->a+=lim x->a-

Answer
This proof is not always true.

If this proof is true on f(x),
then f(x) is said to be continuous at a.

The proof is best done with epsilon, e, and delta, d.
Given that we are looking at lim(x->x0), if this is true,
it can be said no matter how small |x-x0| < e,
a d can be found such that |f(x) - f(x0)| < d.

The limit would vary depending on the function involved.

I will now give two examples, one of which works and one that doesn't work.

Suppose f(x) = (x^2 - 1)/(x - 1).
The lim(x->1)f(x) = 2.

As x goes down to 1, f(x) goes to 2.
Here we have x, f(x):
0.5, 1.5
0.75, 1.75
0.875, 1.875
0.9375, 1.9375
0.96875, 1.96875
0.984375, 1.984375
0.9921875, 1.9921875
0.99609375, 1.99609375
0.998046875, 1.998046875
  
1.5, 2.5
1.25, 2.25
1.125, 2.125
1.0625, 2.0625
1.03125, 2.03125
1.015625, 2.015625
1.0078125, 2.0078125
1.00390625, 2.00390625
1.001953125, 2.001953125

As can be seen in both the limit where x is less than one and
the limit where x is greater than 1, the function value approaches 2.
However, at x = 2, f(x) is undefined.

As a case where this is not true, take f(x) = x/|x| with x0 = 0.
For x > 0, f(x) = 1; for x < 0, f(x) = -1.

Therefore the lim(x->x0) a+ is not the same as lim(x->x0) a-
when a = 0.

Calculus

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