Question

Solve this inequation for x: (e^x-e^-x)/(e^x+e^-x)<m

m is a real number and it is fixed though unknown.

Answer 1

(e^x-e^-x)/(e^x+e^-x)<m can also be expressed as tanh(x)<m (hyperbolic tangent).

e^x-e^-x<me^x+me^-x because e^x+e^-x is always a positive quantity.

e^2x-1<me^2x+m by multiplying by the positive quantity e^x.

e^2x(1-m)<m+1.

We can't divide both sides by 1-m because we don't know that 1-m>0.

-1<tanh(x)<1⇒

(e^x-e^-x)/(e^x+e^-x)≈e^x/e^x=1 for large x>0, so (e^x-e^-x)/(e^x+e^-x)→1 as x→∞.

(e^x-e^-x)/(e^x+e^-x)≈-e^-x/e^-x=-1 for large negative x, so (e^x-e^-x)/(e^x+e^-x)→-1 as x→-∞.

When m≥1, the inequality is always true.

So we now look at |m|<1 (that is, -1<m<1). Under these conditions 1-m>0, so:

e^2x<(1+m)/(1-m), and x<½ln[(1+m)/(1-m)] or x<ln(√[(1+m)/(1-m)]) for -1<m<1.

(e^x-e^-x)/(e^x+e^-x)<m is always true for m≥1, and is never true for m≤-1. (Example: m=0⇒x<0.)