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Continuity and Limits

Evaluating Limits

Problems

Problems

In this text we'll just introduce a few simple techniques for evaluating limits and show you some examples. The more formal ways of finding limits will be left for calculus.

A limit of a function at a certain x -value does not depend on the value of the function for that x . So one technique for evaluating a limit is evaluating a function for many x -values very close to the desired x . For example, f (x) = 3x . What is f (x) ? Let's find the values of f at some x -values near 4 . f (3.99) = 11.97, f (3.9999) = 11.9997, f (4.01) = 12.03, andf (4.0001) = 12.0003 . From this, it is safe to say that as x approaches 4 , f (x) approaches 12 . That is to say, f (x) = 12 .

The technique of evaluating a function for many values of x near the desired value is rather tedious. For certain functions, a much easier technique works: direct substitution. In the problem above, we could have simply evaluated f (4) = 12 , and had our limit with one calculation. Because a limit at a given value of x does not depend on the value of the function at that x -value, direct substitution is a shortcut that does not always work. Often a function is undefined at the desired x -value, and in some functions, the value of f (a)≠ f (x) . So direct substitution is a technique that should be tried with most functions (because it is so quick and easy to do) but always double-checked. It tends to work for the limits of polynomials and trigonometric functions, but is less reliable for functions which are undefined at certain values of x .

The other simple technique for finding a limit involves direct substitution, but requires more creativity. If direct substitution is attempted, but the function is undefined for the given value of x , algebraic techniques for simplifying a function may be used to findan expression of the function for which the value of the function at the desired x is defined. Then direct substitution can be used to find the limit. Such algebraic techniques include factoring and rationalizing the denominator, to name a few. However a function is manipulated so that direct substitution may work, the answer still should be checked by either looking at the graph of the function or evaluating the function for x - values near the desired value. Now we'll look at a few examples of limits.

What is ?

Figure %: f (x) =
By direct substitution and verification from the graph, = - .

What is ?

Figure %: f (x) =
Direct substitution doesn't work, because f is undefined at x = 1 . By facotring the denominator into (x + 1)(x - 1) , though, the (x - 1) term cancels on the top and bottom, and we are left evaluating . By direct substitution, the limit is .

Consider the function f (x) = xforx < 0, f (x) = x + 1forx≥ 0 . What is f (x) , what is f (x) , and what is f (x) ?

Figure %: f (x) = x for x < 0 , f (x) = x + 1 for x≥ 0
The one-sided limit from the left is 0 . This we can tell both from direct substitution and by studying the graph. Using the same techniques, we find the one-sided limit from the right is 1 . By the rules of the nonexistent limit, f (x) does not exist, because is f (x)≠ f (x) .

Consider the function f (x) = xforallx≠3, f (x) = 2forx = 3 . What is f (x) ?

Figure %: f (x) = x for all x≠3 , f (x) = 2 for x = 3
Direct substitution gives the limit at 2 , but more careful inspection of the graph and the values surrounding x = 3 show that in fact the limit of f at x = 3 is 3 . This is a prime example of how the value of a function at x has no impact on the limit of that function at x .

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