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

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#

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)#"