# Is a graph still continuous and differentiable if it has a removable discontinuity? Is a graph still continuous if it has a removable discontinuity?

No. At any point where the function has a discontinuity of any kind, it won't be continuous (by definition of "discontinuous").

Remember that for a function f to be continuous at a, we must have

`lim_(x->a)f(x)=f(a)`.

For an example of a function with a removable discontinuity, take

`f(x)={(x if x!=0),(1 if x=0):}`

This is not continuous at x=0 because

`lim_(x->0)f(x)=0` , but `f(0)=1` , so the limit doesn't equal the function value at that point. This is a removable discontinuity because it can be fixed, if desired, by redefining the single point ` ``f(0)` to be 0.

As far as differentiability, have you learned that if `f` is differentiable at `a`, then `f` must be continuous at `a`? This is logically equivalent to saying that if `f` is *not *continuous at `a`, then `f` is *not *differentiable at `a`.

**So if a function is discontinuous at some point a, then it isn't continuous or differentiable there. This doesn't depend on what type of discontinuity it is.**

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