On-line Math 21

1.1  Examples of basic limits

Example 1 Let f(x) = 3x²+1 . Then

lim x® 2  f(x) = 13.

Solution

You can see this most easily by simply plugging into the expression for f(x) at x = 2 . That ought to be the value of f(2) since f(x) has such a simple expression, and that is the case. The general situation of why and when you can evaluate some limits just by plugging in are explained in [Link: continuity], and [Link: limits-computing]

Example 2 Let

f(x) = ì í î 3x+1 if x ³ 1 x+2 if x < 1 .

Then

lim x® 1  f(x)

does not exist.

The reason it doesn't exist is that, as you approach from the left ( x < 1 ), the values get towards 3, while from the right ( x > 1) , the values get towards 4. This does have what are called one-sided limits (on either side). [Link to one-sided limits]

Example 3 Let

f(x) = ì í î x2 if x ¹ 1 5 if x = 1 .

Then

lim x® 1  f(x) = 1,

even though f(1) = 5 .

Here the function is purposely defined to have the wrong value at 1. The limit

lim x® 1  f(x)

is what f(1) ought to be.

This example stresses the predictability of values of the function. This function is bad in that it doesn't have the expected value at 1. The previous example had no value that f(1) should be; from one side it should be one thing, from the other side it should be another. But f has to have just one value.

Example 4 Let f(x) = sin(1/x) . Then

lim x® 0  f(x)

doesn't exist.

This one looks much worse than Example 2.. Here, as x® 0 , f(x) is wildly switching between -1 and 1.

Example 5 Let

f(x) = x3+1 x2-1 .

Then

lim x® -1  f(x) = -3 2 .

Copyright (c) 2000 by David L. Johnson.