1. ## Singularities or discontinuities

ok I have understand how to do it but i ave confuse when is removable and when is non-removable

here is the question:
Find any singularities or discontinuities in the following functions:

f(x)=(x+3)/(x^2-9)
(an:non-removable singularity at x=3, removable singularity at x=-3, put f(x)=-1/6)

f(x)=cosx/(x-π/2)
(an:removabe singularity at x=π/2, put f(π/2)=-1)

2. A discontinuity is removable if the limit exists

Remember Continuity means lim x->a f(x) = f(a)

which implies
1. f(a) is defined
2. lim x->a f(x)
3. 1 = 2

For example

f(x)=(x+3)/(x^2-9)

limas x->-3 = -1/6 so if we define f(-3) =-1/6 all conditions are met

limasx->3 DNE fails 2 cannot remove discontinuity

3. why the lim x->3= -1/6!???!?i found it -infinite

4. Re read the post I said lim x-> -3 = -1/6

and lim x-> 3 DNE

5. Originally Posted by NaNa
ok I have understand how to do it but i ave confuse when is removable and when is non-removable

here is the question:
Find any singularities or discontinuities in the following functions:

f(x)=(x+3)/(x^2-9)
(an:non-removable singularity at x=3, removable singularity at x=-3, put f(x)=-1/6)

f(x)=cosx/(x-π/2)
(an:removabe singularity at x=π/2, put f(π/2)=-1)
The simple idea is that if $f(x)$ is undefined at a point $x_0$ but the limit at the point $\lim_{x \to x_0}f(x)=c < \infty$ the we can extend the function to the value c at $x_0$ ie $f(x_0)=c$

Edit: Geez I am really late

6. i find the same answer!?
where do you subtract the limit?!

7. I'm not sure what you mean by " subtract the limit" ?

8. Hmm is ok I have understand
thanks a lot!