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QuestionFind the limit of f(x)=sin⁡4x4sin⁡3x+sin⁡(x/3)9xf(x)=\frac{\sin 4x}{4\sin 3x}+\frac{\sin (x/3)}{9x} as x→0x \to 0.

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=sin⁡4x4sin⁡3x+sin⁡(x/3)9xf(x)=\frac{\sin4x}{4\sin3x}+\frac{\sin (x/3)}{9x} . We need to find the limit as xx approaches 00

STEP 2

We will solve this problem by breaking it down into two separate limits and then adding the results together. The first limit is \lim_{x\to0}\frac{\sin4x}{4\sinx} and the second limit is lim⁡x→0sin⁡(x/)9x\lim_{x\to0}\frac{\sin (x/)}{9x}.

STEP 3

We will first solve the limit \lim_{x\to0}\frac{\sinx}{\sin3x}. We can apply the limit property lim⁡x→0sin⁡xx=1\lim_{x\to0}\frac{\sin x}{x}=1 by making the argument of the sine function and the denominator the same.

STEP 4

To make the argument of the sine function and the denominator the same, we multiply and divide the expression by 33.
lim⁡x→0sin⁡4x4sin⁡3x=lim⁡x→03sin⁡4x12sin⁡3x\lim_{x\to0}\frac{\sin4x}{4\sin3x} = \lim_{x\to0}\frac{3\sin4x}{12\sin3x}

STEP 5

Now, we can rewrite the limit as followslim⁡x→03sin⁡4x12sin⁡3x=lim⁡x→0sin⁡4x4x⋅3xsin⁡3x\lim_{x\to0}\frac{3\sin4x}{12\sin3x} = \lim_{x\to0}\frac{\sin4x}{4x} \cdot \frac{3x}{\sin3x}

STEP 6

Apply the limit property lim⁡x→0sin⁡xx=1\lim_{x\to0}\frac{\sin x}{x}=1 to both fractions.
lim⁡x→0sin⁡4x4x⋅3xsin⁡3x=1⋅1=1\lim_{x\to0}\frac{\sin4x}{4x} \cdot \frac{3x}{\sin3x} =1 \cdot1 =1

STEP 7

Now, we will solve the second limit lim⁡x→0sin⁡(x/3)9x\lim_{x\to0}\frac{\sin (x/3)}{9x}. We can apply the limit property lim⁡x→0sin⁡xx=1\lim_{x\to0}\frac{\sin x}{x}=1 by making the argument of the sine function and the denominator the same.

STEP 8

To make the argument of the sine function and the denominator the same, we multiply and divide the expression by 33.
lim⁡x→0sin⁡(x/3)x=lim⁡x→03sin⁡(x/3)27x\lim_{x\to0}\frac{\sin (x/3)}{x} = \lim_{x\to0}\frac{3\sin (x/3)}{27x}

STEP 9

Now, we can rewrite the limit as followslim⁡x→3sin⁡(x/3)27x=lim⁡x→sin⁡(x/3)(x/3)⋅3x9\lim_{x\to}\frac{3\sin (x/3)}{27x} = \lim_{x\to}\frac{\sin (x/3)}{(x/3)} \cdot \frac{3x}{9}

STEP 10

Apply the limit property lim⁡x→0sin⁡xx=\lim_{x\to0}\frac{\sin x}{x}= to the first fraction and simplify the second fraction.
lim⁡x→0sin⁡(x/3)(x/3)⋅3x9=⋅3=3\lim_{x\to0}\frac{\sin (x/3)}{(x/3)} \cdot \frac{3x}{9} = \cdot \frac{}{3} = \frac{}{3}

STEP 11

Now that we have the limits of both parts of the function, we can add them together to get the limit of the entire function.
lim⁡x→0f(x)=+3=43\lim_{x\to0}f(x) = + \frac{}{3} = \frac{4}{3}So, the limit of the function f(x)=sin⁡4x4sin⁡3x+sin⁡(x/3)9xf(x)=\frac{\sin4x}{4\sin3x}+\frac{\sin (x/3)}{9x} as xx approaches 00 is 43\frac{4}{3}.

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