Derivative at a point - AP Calculus AB

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.

Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

, and

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

.

The function is not differentiable at . So the derivative at does not exist.