Now we can see that x, the length of , is less than or equal to the length of . So x tanx.

You have seen how to find the limit of different polynomial and rational functions. Now, let's study the limits of some trigonometric functions.

Recall from the definition of sin x and cos x that the coordinates of the point P on the unit circle are (cos x, sin x).

As x approaches 0, P approaches the point (1,0). So, cos x approaches 1, and sin x approaches 0.

Therefore, and .

Now consider the function . Notice that f(x) is not defined when x = 0. Is defined? If so, what is it?

We can make a guess at the value of this limit by looking at values of the function for positive and negative values of x close to 0.

From the table and the graph, we can guess that .

Next, we will show a geometric proof that this guess is in fact correct.

Assume that x is an angle between 0 and . The figure to the right shows a sector of a circle with center O, central angle x, and radius 1. The line is drawn perpendicular to , the radius of the circle.

By the definition of radian measure, . Also, using the definition of sine, . But is the radius of the circle, which is 1 unit long. So BC = sinx.

From the diagram you can see that BC .

So, sinx x. Dividing both sides by x, we get .

Now, we'll draw the line tangent to the circle at A, and we'll extend to form a triangle. By the definition of tangent, tanx = .

But is the radius of the circle which is 1 unit long. So tanx = AD.

Recall that tanx = , so x .

We selected x to be an angle between 0 and . Therefore, both x and cosx are positive and we can multiply both sides of the inequality by . Doing so, we get cosx .

Earlier we saw that 1, so cosx 1.

As x approaches 0, cos x approaches 1. Therefore, as approaches 0, it is squeezed between 1and 1 and therefore approaches 1. So, .

Let's find .

To find this limit, we need to make the expression more similar to , which we know is equal to 1. So first we multiply the numerator and denominator of by 5.

Using limit laws, this limit is equal to .

Now, we substitute q for 5x. As x approaches zero, 5x approaches zero, so q also approaches zero.

So the limit can be written as .

We know that is 1. So, , or 5. Therefore, is 5.

Evaluate . Click "Submit" when finished.

Copyright 2006 The Regents of the University of California and Monterey Institute for Technology and Education