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?

Solution for Problem 1 >>

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 .

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Problem : Consider the function

f (x) = ,

Solution for Problem 2 >>

Multiply the numerator and denominator of the expression defining f (x) by x ² 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.

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Problem : Compute .

Solution for Problem 3 >>

= = = - 2

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