Re: showing discontinuity

Quote:

Originally Posted by

**Arron** f(x) =

e^x, x<0

cos(pi x) , 0 < or equal x < or equal 1

x^2 - 3/2, x > 1

I need to prove that f is discontinuous at 1.(by manipulating inequalities, rather than using calculus)

I have,

we guess that f is discontinuous at 1, find one {Xn} in A such that

Xn ---> a but f(Xn) does not ---> f(a)

I am not sure what sequence to use, can anyone help?

Well your function is

For a function to be continuous at a point, the function needs to approach the same value from the left as it does from the right.

To the left of , the function is equal to , and so will approach as from the left.

To the right of , the function is equal to , and so will approach as from the right.

Since the function at does not approach the same value from the left as it does from the right, the function is discontinuous at .

Re: showing discontinuity

Thanks. I also need to prove that F is continuous at 0, do I prove this in the same way?

Re: showing discontinuity

Quote:

Originally Posted by

**Arron** Thanks. I also need to prove that F is continuous at 0, do I prove this in the same way?

Yes indeed.