How do I locate discontinuities of functions?

Question

so this is a prime example of what i don't know how to do....

y=1/(1+e^(1/x))

how can i tell what the discontinuities are without using a graphing calculator?

how do i know that e^(1/x)=-1

Answer 1

y=1/(1+e^(1/x))

1/x not exits if x=0 --> D=Cont = |R - {0}

lim{x-->0} 1/(1+e^(1/x)) = 1 if x-->0(-) and 0 if x-->0(+) -->

D. salto finito

Saludos.

Answer 2

You'll get a discontinuity for any value of x (the independent variable) that makes the denominator zero, or if it results in something that is undefined.

In this case e^(1/x) = -1 if denominator is 0
Solving this rsults in Ln(-1) which is undefined. Hence the function has a hole at x = 0 and is thus discontinuous there.

Now the function 1/(1-x^2) has a 0 denominator if x = +/- 1. It is therefore discontinuous at x =1 and x = -1. And it has vertical asymptotes at thes vaues of x, not just a hole.

Answer 3

you will get a discontinuity for any fee of x (the self sufficient variable) that makes the denominator 0, or if it leads to something this is undefined. to that end e^(a million/x) = -a million if denominator is 0 fixing this rsults in Ln(-a million) this is undefined. hence the function has a hollow at x = 0 and is for that reason discontinuous there. Now the function a million/(a million-x^2) has a nil denominator if x = +/- a million. this is hence discontinuous at x =a million and x = -a million. And it has vertical asymptotes at thes vaues of x, no longer in basic terms a hollow.

Answer 4

When you have a zero in the denominator. In this example, when x = 0.