Question

Hey guys, can you please help me out with this =)
Tell me on the graph why the point x=4 is non removable in the following function y=5/(x-4)

All the help will be appreciated! Many thanks

Answer 1

because at x=4 there is a vertical asymptote... no matter how you redefine the function at x=4, the function will never be continuous

Answer 2

perhaps it would be more enlightening to see an example where you can remove the singularity
if \(f(x)=\frac{x^2-16}{x-4}\) then you could remove the discontinuity because if \(x\neq 4\)this is the same as \(f(x)=\frac{(x+4)(x-4)}{x-4}=x+4\) so you could remove it by saying if \(x=4\) then \(f(x)=8\)

Answer 3

but in your example you cannot factor and cancel thereby removing the discontinuity.
that is why it is not "removable'

Answer 4

Thankyou everyone :)