1. ## showing discontinuity

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?

2. ## Re: showing discontinuity

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?

$\displaystyle \displaystyle f(x) = \begin{cases}e^x\textrm{ if }x < 0 \\ \cos{(\pi x)} \textrm{ if } 0 \leq x \leq 1 \\ x^2 - \frac{3}{2} \textrm{ if }x > 1\end{cases}$

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 $\displaystyle \displaystyle x = 1$, the function is equal to $\displaystyle \displaystyle \cos{(\pi x)}$, and so will approach $\displaystyle \displaystyle \cos{(\pi \cdot 1)} = -1$ as $\displaystyle \displaystyle x \to 1$ from the left.

To the right of $\displaystyle \displaystyle x = 1$, the function is equal to $\displaystyle \displaystyle x^2 - \frac{3}{2}$, and so will approach $\displaystyle \displaystyle 1^2 - \frac{3}{2} = -\frac{1}{2}$ as $\displaystyle \displaystyle x \to 1$ from the right.

Since the function at $\displaystyle \displaystyle x = 1$ does not approach the same value from the left as it does from the right, the function is discontinuous at $\displaystyle \displaystyle x = 1$.

3. ## Re: showing discontinuity

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

4. ## Re: showing discontinuity

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.