We see that if a function is differentiable at a point, then it must be continuous at that point.
Assuming that exists, we want to show that is continuous at , hence we must show that Starting with we multiply and divide by to get
Since we apply the Difference Law to the left hand side and use continuity of a constant to obtain that Next, we add on both sides and get that Now we see that , and so is continuous at .
This theorem is often written as its contrapositive:
If is not continuous at , then is not differentiable at .
Thus from the theorem above, we see that all differentiable functions on are continuous on . Nevertheless there are continuous functions on that are not differentiable on .
So for the function to be continuous, we must have We also must ensure that the function is differentiable at . In other words, we have to ensure that the following limit exists
In order to compute this limit, we have to compute the two one-sided limits and since changes expression at . Write with me
Hence, we must have Ah! So now the equation that must be satisfied
Therefore, . Thus setting and will give us a function that is differentiable (and hence continuous) at .
Can we tell from its graph whether the function is differentiable or not at a point ?
Each of the figures A-D depicts a function that is not differentiable at .
In figures – the functions are continuous at , but in each case the limit does not exist, for a different reason.
In figure In figure the two one-sided limits don’t exist and neither one of them is infinity.
So, if at the point a function either has a ”jump” in the graph, or a corner, or what looks like a “vertical tangent line”, or if it rapidly oscillates near , then the function is not differentiable at .