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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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