how do you find the limit
lim as x->0 sin4x/sin6x
2006-09-19 10:51:14 · 3 answers · asked by Lauren 1 in Science & Mathematics ➔ Mathematics
Calculus books differ on when they introduce L'Hopital's Rule. If you have covered this rule in class, then it's available for use in this problem. (L'Hopital's Rule is used when direct substitution of limits yields indeterminant fractions. This example would give 0/0 if 0 is directly substituted.)
L'Hopital's Rules says to take the derivative of the numerator and the derivative of the denominator _seperately_ (which is different from using the Quotient Rule):
lim as x->0 of sin(4x) / sin(6x)
= lim as x->0 of 4cos(4x) / 6cos(6x)
Now direct substitution yields
= 4 (cos 0) / 6 (cos 0)
= 4/6
= 2/3.
2006-09-19 11:08:16 · answer #1 · answered by HiwM 3 · 0⤊ 1⤋
limit x>0 {sin4x/4x}/{sin6x/6x}*4/6
=2/3
2006-09-19 10:57:15 · answer #2 · answered by raj 7 · 2⤊ 1⤋
There is a rule lim(x->0)sin(x)/x = 1
It is the same for lim(x->0) sin(kx)/(kx) = 1
Now you try to put your function in this form, without changing the value.
lim(x->0) sin(4x)/sin(6x) = lim(x->0) (4sin(4x)/4x) / (6sin(6x)/6x)=
4.1/6.1 = 4/6
2006-09-19 11:45:58 · answer #3 · answered by vahucel 6 · 1⤊ 0⤋