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Problem :
Define a function by f (x) = x - 3 for x≠1 and let f (1) = 1. At which values of x
is f (x) continuous? Does f (x) have a limit at x = 1? If so, what is this limit?
A graph of this function is displayed below.
Figure %: Plot of f (x) = x - 3 for x≠1 and let f (1) = 1
As x approaches 1, the values of f (x) approach 1 - 3 = - 2, so
f (x) = - 2.
However, f (1) = 1≠ - 2. Therefore, f (x) is not continuous at x = 1. It is clear from
the graph that f (x) is continuous at all other values of x.
Problem :
Consider the function
f (x) = ,
defined for x≠ - 1, 0. Does f (x) have a limit at x = 0? If so, what is the limit?
Multiply the numerator and denominator of the expression defining f (x) by
x2 to obtain
g(x) =
which will be equal to f (x) for all x≠ 0. This new function clearly has limit 2 at
x = 0, thus so does f (x). It does not make sense to ask whether f (x) is continuous at
0, because 0 is not in its domain.
Problem :
Compute .
= = = - 2
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