Limits at Corners
This question was posted by an anonymous reader. Do functions with corners have limits (at the corner)? For example, the function:
With "corners"? What is that? Let's graph this function to understand what we mean by corners:
This function has a corner at x = 0. Does the limit exist at that point? The answer is yes, and this can be understood just using the intuition of what is the limit
. The limit is what value the function approaches when x (independent variable) approaches a point.
In this case, to make x close to 0 we can make it positive or negative. In these cases is when the concept of lateral limits comes in handy. If x
takes only positive values and approaches 0 (approaches from the right), we see that f(x) also approaches 0.
Likewise, if x approaches 0 with negative values (from the left), f(x) approaches 0. So both lateral limits are equal to 0. This implies that the limit
itself is zero! Why? Because of our understanding of limit. No matter how x approaches 0, f(x) approaches 0 when x is near 0.
The same argument may be used to prove the limit of other functions with corners similar to this one.
Another interesting question about functions like this one is whether or not the derivative exists at the corner. The answer is no. Derivatives do not
exist at corner points. This is explained, once again, using an intuitive idea.
The geometric interpretation of the derivative
is that it is the slope of the tangent line at that point. The existence of the derivative is therefore
related to the existence of a tangent line. But just imagine, how many tangent lines can you draw at point x=0? Infinite! There isn't "a" tangent line.
So, the derivative does not exist at that point. Return to Calculus Questions