Let f(x) =e^x - e^-x & g(x)= e^x + e^-x .Also, let h(x)= f(x)/g(x). Then... cont.?

Question

lim h(x) =
x=>-oo

Answer 1

Aegor is correct, but here's another way to see it. Simplify h(x) by multiplying top and bottom of the fraction by e^x; then
h(x) = [e^(2x) - 1]/[e^(2x) + 1]. (Note careful use of brackets.) Clearly, as x tends to minus infinity, e^(2x) tends to zero, so h(x) tends to just -1/1, i.e. -1.

Live long and prosper.

Answer 2

I was cheap and used my calc on this one. It appears that as x approaches negative infinity, the function approaches -1, and as x approaches positive infinity, the function approaches 1.

However, here is something: f(x) is an odd function, and g(x) is an even function. They have the same value when x approaches positive infinity, but their values become equal but f(x) has the negative value of g(x) as x approaches negative infinity. Thus, f(x)/g(x) as x approaches negative infinity becomes -g(x)/g(x) = -1.


Hope this helps.