Question

Does the limit `lim_(x->3) (f(x)-f(3))/(x-3)` always exist?

Answer 1

Kindly refer to the Discussion given in the Explanation.

Explanation:

The Limit under reference may or may not exist.

Its existence depends upon the definition of the function `f.`

Consider the following Examples :

`(1) : f(x)=x, x in RR.`

Clearly, the Limit =`lim_(x to 3) {f(x)-f(3)}/(x-3),`

`=lim_(x to 3)(x-3)/(x-3)............[because, f(3)=3]`

`=lim_(x to 3) 1.`

We find that, `lim_(x to 3) {f(x)-f(3)}/(x-3),` exists, and, is `1.`

`(2) : f(x)=|x-3|, x in RR.`

Remember that,

`AA x in RR, |x|=x; if x>=0, &, |x|=-x, if x < 0.`

Since, `f(3)=|3-3|=0,` we have, `{f(x)-f(3)}/(x-3)=|x-3|/(x-3).`

`"Now, as "x to 3-, x < 3 :. (x-3) <0.`

`:. |x-3|=-(x-3).`

`:. lim_(x to 3-){f(x)-f(3)}/(x-3),`

`=lim_(x to 3-) {-(x-3)}/(x-3),`

`rArr lim_(x to 3-) {f(x)-f(3)}/(x-3)=-1....................(star^1).`

On the other hand, as `x to 3+, x>3. :. |x-3|=(x-3).`

`rArr lim_(x to 3+){f(x)-f(3)}/(x-3)=1..........................(star^2).`

`(star^1), &, (star^2),`

`rArr lim_(x to 3-) {f(x)-f(3)}/(x-3)=-1!=1=lim_(x to 3+){f(x)-f(3)}/(x-3).`

We conclude that, `lim_(x to 3){f(x)-f(3)}/(x-3)` does not exist.

Enjoy Maths.!

Answer 2

Recall the limit definition of the derivativbe, that is:

`f'(a) = lim_(x rarr a) (f(x)-f(a))/(x-a)`

We have:

`L = lim_(x->3) (f(x)-f(3))/(x-3)`

And so clearly:

`L = f'(3)`

Without further knowledge of the function we cannot determine if the limits exist. If it were known that `f(x)` was differentiable over some domain that included `x=3` then we could conclude that the limit exists.