Please help.?

Question

Consider the function f(x)=2/(1+e^(1/x)). Now why does the graph have a horizontal asymptote at y=1 and why the function has a nonremovable discontinuity at x=0??????

Answer 1

to see the first part, think about the limit x=>+inf then 1/x=>0 so that e^(1/x)=>1 and f(x)=>2/2 = 1 (from above)

you can get a similar result for x=>-inf except that the function approaches 1 from below. Why?

now consider x=>0 from above, then 1/x=>+inf
x=>0 from below then 1/x =>-inf
so that the function is discontinuous at x=0.

to see that it's non-removable, think of how you would transform the function 2/[1+e^(1/x)] to eliminate the dependency on 1/x near zero. The mixed linear and transcendental functions will probably stymie you.

hth

f(x) = 2[1-e^(-1/x)] / sinh(1/x)

Answer 2

when x -> infinity e^(1/x) -> 0 so f(x)-> 2/1 = 2

when x = 0, 1/x is undefined so e^(1/x) is also undefined.