is the limit as x approaches 0 of sin3x over 3x equal to zero?
thank you!
do you mean?
(sin3x)/3x
hint: L'Hopital's Rule
take the derivative of the numerator and denominator
& then plug in zero
sorry--
basically
lim [sin 3x / 4x)
x-> 0
i multiplied& eventually got to
.75* lim (sin 3x / 3x)
x-> 0
so i figured since (lim (sinx/x)
x-> 0
was equal to zero, then
lim (sin3x/ 3x) also equaled 0
x-> 0
is that right? thank you
take the derivative of the numerator and denominator individually.
then, plug in zero for x.
sin3x
hint: chain rule
To find the limit as x approaches 0 of sin(3x)/3x, we can start by evaluating this expression directly. Substituting x = 0 into the expression gives sin(3(0))/3(0), which is 0/0. This is an indeterminate form, meaning we can't determine the limit by simple substitution.
To further analyze this expression and find the limit, we can use L'Hôpital's Rule. L'Hôpital's Rule states that if we have an indeterminate form, we can differentiate the numerator and the denominator with respect to x and take the limit again. The result will give us the same limit as the original expression.
Let's differentiate both the numerator and denominator:
The derivative of sin(3x) is cos(3x) * d(3x)/dx = 3cos(3x)
The derivative of 3x is 3
Now we can rewrite the expression in terms of the derivatives:
lim x->0 [sin(3x)/3x] = lim x->0 [(3cos(3x))/(3)]
Notice that the 3s can cancel out, so we have:
lim x->0 cos(3x)
Now we can substitute x = 0 into the expression. cos(3(0)) = cos(0) = 1. Therefore,
lim x->0 sin(3x)/3x = 1
The limit as x approaches 0 of sin(3x)/3x is equal to 1, not zero.