lim as x approaches 0
(e^4x-1)/sin(2x)
assuming you mean:
(e^(4x) - 1)/sin(2x)
= lim (4 e^(4x))/(2cos(2x)) as x ---> 0
= 4e^0 / 2cos0
= 4(1)/2 = 2
lim as x approaches 0
(e^4x-1)/sin(2x)
I know you use L-hopitals rule and I know the answer is 2 but I keep getting 1. Can someone help
2 answers
Thank you that makes sense!