lim_(theta->0)(cos(6 theta)tan(6 theta))/(theta)
lim_(theta->0)(cos(6 theta)tan(6 theta))/(theta)
=Lim (θ→0)[cos(6θ)sin(6θ)/cos(6θ)]/θ
=Lim [6 (sin(6θ)/6θ)]
=Lim [6*1]
=6
Note:
Lim sin(x)/x =1
x→0
To evaluate the limit, let's simplify the expression first.
Taking the limit as theta approaches 0, we have:
lim(theta->0) [cos(6theta) * tan(6theta)] / theta
Since the numerator contains both cos(6theta) and tan(6theta), which are both trigonometric functions, we can rewrite tan(6theta) as sin(6theta)/cos(6theta) using the identity tan(x) = sin(x)/cos(x). This gives us:
lim(theta->0) [cos(6theta) * (sin(6theta) / cos(6theta))] / theta
Next, we can cancel out the common factors of cos(6theta) in the numerator and denominator:
lim(theta->0) sin(6theta) / theta
Now, this expression appears to be an indeterminate form of the type 0/0 because both sin(6theta) and theta approach 0 as theta approaches 0. To evaluate this limit, we need to apply L'Hôpital's rule.
L'Hôpital's rule states that if we have an indeterminate form 0/0 (or ∞/∞), we can differentiate the numerator and the denominator until we get a determinate form.
Taking the derivative of both the numerator and the denominator with respect to theta, we get:
lim(theta->0) [6cos(6theta)] / 1
Now we can evaluate the limit directly, without any indeterminate forms:
lim(theta->0) 6cos(6theta) / 1
Since we are evaluating the limit as theta approaches 0, we substitute theta = 0 into the expression:
6cos(6(0)) / 1
Simplifying further:
6cos(0) / 1
The cosine of 0 is equal to 1, so the final answer is:
6(1) / 1 = 6
Therefore, the limit as theta approaches 0 of (cos(6theta)tan(6theta))/theta is equal to 6.