Question

If `f(x) = { (x^2, x !=2), (2, x=2) :}` then evaluate `lim_(x rarr 2) f(x)`?

Answer 1

`lim_(x rarr 2) f(x) = 4`

Explanation:

In order to evaluate a limit we are not interested in the value of the function at the limit, just the behaviour of the function around the limit:

We have:

`f(x) = { (x^2, x !=2), (2, x=2) :}`

If we graphed the function would be the parabola `y=x^2` with a hole at `x=2`.

If we just examine the behaviour close to `x=2`, e.g `x=2+-0.001` we get:

`f(2-0.001) = f(1.999) = 3.996001`
`f(2+0.001) = f(2.001) = 4.004001`
`f(2) = 2`

Which would certainly "suggest" that `lim_(x rarr 2) f(x)` is `4` rather than `2`. We can show thi is the case analytically as follows by studying the limit either side of `x=2`:

The Left Handed limit:

`lim_(x rarr 2^-) f(x) = lim_(x rarr 2^-) x^2`
`" " = 2^2`
`" " = 4`

And The Right Handed limit:

`lim_(x rarr 2^+) f(x) = lim_(x rarr 2^+) x^2`
`" " = 2^2`
`" " = 4`

And as both limits are identical we have:

`lim_(x rarr 2^) f(x) = 4`

In both cases we chose the "`x^2`" variant of the function as `x rarr 2` but `x != 2`

Answer 2

Please see the discussion below.

Explanation:

One way to think about the limit of a function as `x` approaches `2` is that it is the value you would expect if you looked at the graph but covered up the graph where `x = 2`.

The limit wants to know "for values of `x` near `2`, is the value of `f(x)` (think of it as `y`) close to just one number. And if so, what number is that?

The limit, `lim_(xrarr2)f(x)` doesn't care what happens when `x = 2`, just what happens when `x` is close to `2`

The question of what happens whar `x` equals two is the question of the value of `f(2)`. In your example, `f(2) = 4` but for `x` near `2` (and not equal to `2`) `f(x) = x^2` is close to `4`. So that is the limit.

Finally, functions like the one in this question (kind of strange, with weird points) are very important to learning the difference between:

"Find `f(a)`" and "Find `lim_(xrarra)f(x)`"