In the last section, we discussed continuity which is very important in Calculus. In this section we are going to cover the limits of some trigonometric functions namely, the sine and cosine functions as listed below-

These are special limits because even though we are dividing by zero in the denominator the limit is still defined. Let's try some problems involving limits of trig functions.

Suppose we are asked to evaluate the following limit-

First notice that this limit is tending to zero from the positive side. Why? There is a square root in the denominator, and taking a square root of a negative number leads to a complex number which is beyond the scope of the calculus being discussed here. So for all real numbers, the square root only works for positive numbers or zero.

Ok so we are taking a limit of a function from the right. What's next? The function is not quite in the form of the above trigonometric limit because of that square root. Let's multiply the numerator and denominator by the square root of sin x:

and we are left with:

Now this limit looks like: and we can now finish our problem to:

Let's try a problem with a cosine:

You might be wondering where to start from. We can't evaluate the limit the way it is now because we would be dividing by zero in the denominator. We can begin by factoring out a cosine from the numerator. So now we have:

We sort of have the limit looking like: , but the cosine and one are reversed. We can factor out a negative one. So now our limit looks like:

We can now replace (1 - cos x)/x with 0 since that is what the limit evaluates to. The limit now looks like:

We're not dividing by 0 in the denominator so now we can evaluate the limit as it is, which turns out to be 0.