# A Confusing Question to Me: Solving Limits by Continuity

by Cody
(San Marcos)

When we're told to solve limits, the first thing we must do is to try to evaluate the function at the point we taking the limit. In this question we have the limit: All of these functions are continuous functions. Usually, the only case where you'll encounter a discontinuous function in these types of problems is when the function is not defined, because the denominator becomes zero somewhere, or we take the square root of a negative number.

In this case, if we evaluate the function at the point we're taking the limit (in this case 0), we get: Now, we just need to remember the definition of secant from trigonometry, and the fact that cos(0)=1: Since these functions are continuous, and this particular function is defined at the point, this value must be the limit. Why? Because one definition of a continuous function is that: This of course is only valid for values "a" in the domain of the function.

So, applying this with this particular function we have that: 