NEED HELP WITH LIMITS (AP calculus) x approaching some number from the negative side and from the positive side??

Limit as x approaches -2 and 2 from the negative and positive side.

x approaches -2 from the negative side = -1/0 so is the zero a little positive or a little negative??

1 Answer
Dec 17, 2017

Please see below.

Explanation:

Consider lim_(xrarr-2^-)(x+1)/(x+2)

As xrarr-2, the numerator approaches -1 and the denominator approaches 0.

As xrarr-2 from the left, that is, as xrarr-2^- we have x < -2 so x+2 < -2+2 = 0.
That is: the denominator approaches 0 through negative values. (negative fractions if you like.)
I like to use the notation: lim_(xrarr-2^-)(x+2) = 0^- to indicate the direction from which 0 is being approached.
(I am NOT making 0 a negative number. Just indicating which side the value are on.)

The numerator is going to -1 and the denominator is approaching 0 through negative values, so the quotient is increasing without bound.

It is a bit tedious to write the explanation:

lim_(xrarr-2^-)(x+1)/(x+2) does not exist because as x approaches -2 from the left, the values of the quotient increase without bound.

It is simpler (though misleading) to write

lim_(xrarr-2^-)(x+1)/(x+2) = oo

On the other hand
Conside lim_(xrarr-2^-)(x+1)/(x+2)^2

As xrarr-2, again the numerator goes to -1 and the denominator goes to 0. But this time the denominator is squared, so the values are positive.

The numerator goes to -1 and the denominator goes to 0 through positive values, therefore the quotient is a big negative number

lim_(xrarr-2^-)(x+1)/(x+2) does not exist because as x approaches -2 from the left, the values of the quotient decrease without bound.

We write:
lim_(xrarr-2^-)(x+1)/(x+2) = -oo