Question

How do you find `\lim _ { x \rightarrow 2^ { - } } \frac { f ( x ) - f ( 2) } { x - 2}`?

Answer 1

There is not enough information to determine the limit.

Explanation:

Without knowing what the function is left of `2` and at `2`, we cannot give an answer. It could be anything or it might not exist.

Examples
Example 1
`f(x) = x^2`

`lim_(xrarr2^-) (f(x)-f(2))/(x-2) = lim_(xrarr2^-)(x^2-4)/(x-2) = lim_(xrarr2^-) (x+2) = 4`

Example 2

`f(x) = mx+b` for `m,b` constant

`lim_(xrarr2^-) (f(x)-f(2))/(x-2) = lim_(xrarr2^-)((mx+b)-(2m+b))/(x-2) = lim_(xrarr2^-) (m(x-2))/(x-2) = m`

Example 3

`f(x) = abs(x-2)`

`lim_(xrarr2^-) (f(x)-f(2))/(x-2) = lim_(xrarr2^-)(abs(x-2)-0)/(x-2) = lim_(xrarr2^-) (abs(x-2))/(x-2) = lim_(xrarr2^-)(-(x-2))/(x-2) = -1`

Example 4

`f(x) = {(x+3,"if",x < 2),(x,"if",x >= 2):}`

`lim_(xrarr2^-) (f(x)-f(2))/(x-2) = lim_(xrarr2^-)(x+3-2)/(x-2) = lim_(xrarr2^-) (x+1)/(x-2)` Does Not Exist because as `xrarr2^-`, the quotient decreases without bound.

(We write `lim_(xrarr2^-) (x+1)/(x-2) = -oo`)

These examples do not include all possibilities, but they should give an idea of why the information given is insufficient.