Identifying type of discontinuity

Question

if g(x) = 1/ [e^1/x - 1] what type of discontinuity is at x=0. 

I graphed the function using desmos and I know the function shows jump discontinuity at x=0

but I do not know how to arrive at this answer algebraically or without a graphing calculator. 

Answer 1

To identify continuity (or discontinuity ) of a function f(x) at a point a, you need to find

lim_x→a⁻ f(x), the limit from the left of a

lim_x→a⁺ f(x), the limit from the right of a

f(a), the value of the function at a.

 

Continuity

lim_x→a⁻ f(x) = lim_x→a⁺ f(x) ⇒ lim_x→a f(x) exists

lim_x→a f(x) = f(a)

 

Removable Discontinuity

lim_x→a⁻ f(x) = lim_x→a⁺ f(x) ⇒ limx→a f(x) exists
limx→a f(x) ≠ f(a)

 

Jump Discontinuity

lim_x→a⁻ f(x) ≠ lim_x→a⁺ f(x) ⇒ limx→a f(x) does not exist

 

For this problem

 

lim_x→0− g(x) = lim_x→0⁻ 1/(e^1/x − 1) = 1/[e^1/0− − 1] = 1/(e^−∞ − 1) = − 1

lim_x→0⁺ g(x) = lim_x→0⁺ 1/(e^1/x − 1) = 1/[e^1/0+ − 1] = 1/(e+∞ − 1) = 0

 

Since lim_x→0⁻ g(x) ≠ lim_x→0⁺ g(x), g(x) has a jump discontinuity at 0.