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 .

In order to compute this limit, we have to compute the two one-sided limits and since changes expression at . Write with me

and 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 .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 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 .