lim((g(x)*sin(2x)+e^(2*x)-1)/x),x->0,lim g(x)=5,x->0
So we have
lim[(5*sin(2x) + e^(2x) - 1) /x] , x ⇒ 0 using L'Hospital's Rule we have
lim [(10cos(2x) + 2e^(2x)] / 1. x⇒ 0 =
10*cos(0) + 2e^(0) = 10 + 2 = 12
I have not yet learned L'Hospitals rule.Is there another method for this?
