Calculus Homework Help!! Please!?

Question

Determine the limit of the trigonometric function:

Limit as x->0: cos(6*theta)tan(6*theta) / theta

I have no idea. Please explain how to get the answer! Thanks In Advance!

Answer 1

That equals (sin(6theta))/theta

Plugging in 0 gives 0/0 so there must be a finite limit. I'm gonna assume you don't know L'Hopital's Rule.

I'd say the easiest way to calculate it would be to graph it. Using L'Hopital's rule the answer is:

(6cos(6*0))/1=6cos(0)=6

So you can check the graph and see if it is 6.

Let me know if you have questions.

Answer 2

it would be "cos(6*theta)tan(6*theta) / theta"  unless you meant for x to be theta, in which case it would be pi/30 (make tan sin/cos, and the cos's cancel out)

Answer 3

Assuming that your theta is x (measured in radians), the answer is 6.

Basis:

(1) First, as a slight simplification,

note that cos(6x)*tan(6x) = sin(6x)

(2) So we want to find the limit of [sin(6x)/x] as x nears 0

(3) It is well-known that sin(y) approaches y, as the radian

argument y nears 0, so sin(6x) nears 6x as 6x nears 0.

(4) Therefore, as x becomes very small, [sin(6x)/x] approaches

6x/x, which = 6.

Note: There are other ways to obtain this same result, for example using L'Hospital"s Rule which involves taking derivatives, but the above is probably the simplest method for this case. Enjoy!

Answer 4

this equals cos6* sin6* /cos6* / *   where * =theta

The cos6* cancels out

sin6*/*  and this is the special limit which equals one

The answer is one.

ANSWER

Answer 5

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