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.