Find the limit as x approaches 0 of (tan6x)/(sin2x)

use the good old L'Hospital's Rule:

6sec^2(6x) / 2cos(2x)
6/2 = 3

To find the limit as x approaches 0 of (tan6x)/(sin2x), we can use some properties of trigonometric functions.

First, let's simplify the expression.

Recall that the tangent function can be expressed as sin(x)/cos(x). Therefore, we can rewrite the expression as follows:

(tan6x)/(sin2x) = (sin(6x)/cos(6x))/(sin(2x)).

Now, we can simplify further by canceling out the common factor of sin(x) in the numerator and denominator:

(sin(6x)/cos(6x))/(sin(2x)) = (sin(6x))/(cos(6x)*sin(2x)).

Next, we need to use some trigonometric identities to simplify the expression even more.

The double-angle identity for sine states that sin(2x) = 2sin(x)cos(x).

We can rewrite the expression again using this identity:

(sin(6x))/(cos(6x)*sin(2x)) = (sin(6x))/(cos(6x)*(2sin(x)cos(x))).

Now, we have simplified the expression significantly. However, we still have cos(6x) in the denominator, which could be problematic when x approaches 0.

To handle this, we can use another trigonometric identity, the product-to-sum identity, which states that cos(a)cos(b) = (1/2)[cos(a-b) + cos(a+b)].

Using this identity, we can rewrite the expression once again:

(sin(6x))/(cos(6x)*(2sin(x)cos(x))) = (sin(6x))/(0.5[cos(6x-2x) + cos(6x+2x)]*2sin(x)cos(x)).

Now, we can simplify the expression further:

= (sin(6x))/(sin(x)cos(x)[cos(4x) + cos(8x)]).

At this point, we have eliminated the problematic denominator, and we can proceed to find the limit.

To find the limit as x approaches 0, we substitute x = 0 into the expression:

lim(x->0) (sin(6x))/(sin(x)cos(x)[cos(4x) + cos(8x)]).

Substituting x = 0 into the expression, we get:

lim(x->0) (sin(6*0))/(sin(0)cos(0)[cos(4*0) + cos(8*0)]).

Which simplifies to:

lim(x->0) (0)/(0*cos(0)[1 + 1]).

Notice that we have 0 in the numerator and denominator. This indicates that we have an indeterminate form, and we need further algebraic manipulation to evaluate the limit.

By applying L'Hôpital's rule, we can differentiate the numerator and denominator with respect to x and evaluate the limit of the derivatives:

lim(x->0) [6cos(6x)]/[cos(x)cos(4x) + cos(8x)sin(x) - sin(x)cos(4x)].

By substituting x = 0 into the expression, we can reduce it to:

[6cos(6*0)]/[cos(0)cos(4*0) + cos(8*0)sin(0) - sin(0)cos(4*0)].

Which further simplifies to:

6/[1 + 1 - 0].

And ultimately evaluates to:

6/2 = 3.

Therefore, the limit as x approaches 0 of (tan6x)/(sin2x) is 3.