Question

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??

Answer 1

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`