Math

QuestionFind the limit: limx0sin(6x)sin(2x)\lim _{x \rightarrow 0} \frac{\sin (6 x)}{\sin (2 x)}.

Studdy Solution

STEP 1

Assumptions1. We are asked to find the limit as x approaches0 for the function sin(6x)sin(x)\frac{\sin (6 x)}{\sin ( x)}. . We will use the limit properties and trigonometric identities to solve this problem.

STEP 2

We notice that the limit is of the form 00\frac{0}{0} as x approaches0. This is an indeterminate form, so we can apply L'Hopital's rule, which states that the limit of a ratio of two functions as x approaches a certain value is equal to the limit of the ratios of their derivatives.
limx0sin(6x)sin(2x)=limx0(6cos(6x))(2cos(2x))\lim{x \rightarrow0} \frac{\sin (6 x)}{\sin (2 x)} = \lim{x \rightarrow0} \frac{(6 \cos (6 x))'}{(2 \cos (2 x))'}

STEP 3

Now, we calculate the derivatives of sin(6x)\sin (6 x) and sin(2x)\sin (2 x) using the chain rule. The derivative of sin(u)\sin(u) is cos(u)u\cos(u) \cdot u', where uu' is the derivative of uu.
For sin(6x)\sin (6 x), u=6xu =6x and u=6u' =6. So, the derivative is 6cos(6x)6 \cos (6 x).
For sin(2x)\sin (2 x), u=2xu =2x and u=2u' =2. So, the derivative is 2cos(2x)2 \cos (2 x).

STEP 4

Substitute these derivatives back into the limit.
limx06cos(6x)2cos(2x)\lim{x \rightarrow0} \frac{6 \cos (6 x)}{2 \cos (2 x)}

STEP 5

implify the limit by cancelling out the common factor of2.
limx03cos(x)cos(2x)\lim{x \rightarrow0} \frac{3 \cos ( x)}{\cos (2 x)}

STEP 6

Now, we evaluate the limit as x approaches0.
3cos(60)cos(20)=3cos(0)cos(0)\frac{3 \cos (6 \cdot0)}{\cos (2 \cdot0)} = \frac{3 \cos (0)}{\cos (0)}

STEP 7

Since cos(0)=1\cos(0) =1, the limit simplifies to311=3\frac{3 \cdot1}{1} =3So, the limit of the function as x approaches0 is3.

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