lim
x=>0
2006-07-12 07:15:08 · 5 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
Use the property that
lim [sin(x)/x] = 1
x->0
Multiply the top and bottom by 2
2*sin(14x)/(14x)
= 2(1)
= 2
(lim as x->0 of sin(14x)/(14x) = 1)
2006-07-12 07:19:14 · answer #1 · answered by MsMath 7 · 8⤊ 1⤋
I too would go with L'Hopital's rule (not L'Hospital, haha), and I conclude that the answer is 2.
To use L'Hopitale's rule, you take the derivative of the top and the bottom. The derivative of the top is 14 cos(14x) and the derivative of the bottom is 7.
[14cos(14x)]/7 =
2cos(14x) =
(substitute 0 for x, since the limit approaches 0)
2 cos (14 x 0) =
2 cos (0) =
2 x 1 = 2
There you have it!
2006-07-12 07:39:19 · answer #2 · answered by Tarah 2 · 0⤊ 0⤋
4
2006-07-12 07:17:15 · answer #3 · answered by Grin Reeper 5 · 0⤊ 0⤋
[(sin14x)/7x
lim
x=>0 [(sin2*7x)\7x
=lim
x=>0
=2*1=2 {because (sinax)/ax=1
2006-07-12 07:38:19 · answer #4 · answered by vijeta s 1 · 0⤊ 0⤋
L'Hospital's Rule is the way to go.
2006-07-12 07:20:19 · answer #5 · answered by DoctaB01 2 · 0⤊ 0⤋