Evaluate the following limits :
`lim_(x to oo) (2 sin x - sin 2 x)/(x^(3))`

Answer

To evaluate the limit

\[ \lim_{x \to \infty} \frac{2 \sin x - \sin 2x}{x^3}, \]

we can follow these steps:

### Step 1: Rewrite \(\sin 2x\)
We know that \(\sin 2x = 2 \sin x \cos x\). Therefore, we can rewrite the expression in the limit:

\[ \lim_{x \to \infty} \frac{2 \sin x - \sin 2x}{x^3} = \lim_{x \to \infty} \frac{2 \sin x - 2 \sin x \cos x}{x^3}. \]

### Step 2: Factor out \(2 \sin x\)
Now, we can factor out \(2 \sin x\) from the numerator:

\[ = \lim_{x \to \infty} \frac{2 \sin x (1 - \cos x)}{x^3}. \]

### Step 3: Rewrite \(x^3\)
We can express \(x^3\) as \(x^2 \cdot x\):

\[ = \lim_{x \to \infty} \frac{2 \sin x (1 - \cos x)}{x^2 \cdot x}. \]

### Step 4: Use the identity \(1 - \cos x\)
Using the identity \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\):

\[ = \lim_{x \to \infty} \frac{2 \sin x \cdot 2 \sin^2\left(\frac{x}{2}\right)}{x^2 \cdot x} = \lim_{x \to \infty} \frac{4 \sin x \sin^2\left(\frac{x}{2}\right)}{x^3}. \]

### Step 5: Analyze the limit
As \(x\) approaches infinity, \(\sin x\) oscillates between -1 and 1, and \(\sin^2\left(\frac{x}{2}\right)\) also oscillates between 0 and 1. Therefore, the numerator is bounded while the denominator \(x^3\) approaches infinity.

Thus, we can conclude:

\[ \lim_{x \to \infty} \frac{4 \sin x \sin^2\left(\frac{x}{2}\right)}{x^3} = 0. \]

### Final Answer
Therefore, the limit is:

\[ \lim_{x \to \infty} \frac{2 \sin x - \sin 2x}{x^3} = 0. \]

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