Working on my calc bc homework and mildly stumped by this problem. I know that when graphed the answer is 3, but I need to get it into a form where I can use L'Hopital's rule and work it out.&nbs...
Working on my calc bc homework and mildly stumped by this problem. I know that when graphed the answer is 3, but I need to get it into a form where I can use L'Hopital's rule and work it out.&nbs...
lim 1-cos(3x)/x^2 x→0 I took the derivative and got 3sin(3x)/2x, when I plug in zero I get 0/0 I took the derivative a second time and got 9cos(3x)/2, plugged in...
use L'Hopital rule to find limit.
limx=0(sinh(2x)) (cosh(3x)) find the limit using l'hopital's rule