(ex-e-x)/(ex+e-x)<m can also be expressed as tanh(x)<m (hyperbolic tangent).
ex-e-x<mex+me-x because ex+e-x is always a positive quantity.
e2x-1<me2x+m by multiplying by the positive quantity ex.
e2x(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⇒
(ex-e-x)/(ex+e-x)≈ex/ex=1 for large x>0, so (ex-e-x)/(ex+e-x)→1 as x→∞.
(ex-e-x)/(ex+e-x)≈-e-x/e-x=-1 for large negative x, so (ex-e-x)/(ex+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:
e2x<(1+m)/(1-m), and x<½ln[(1+m)/(1-m)] or x<ln(√[(1+m)/(1-m)]) for -1<m<1.
(ex-e-x)/(ex+e-x)<m is always true for m≥1, and is never true for m≤-1. (Example: m=0⇒x<0.)