Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$
✓
Answer
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}-\sin3\text{x}}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}\big(3\sin\text{x}-4\sin^3\text{x}\big)}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{4\sin^3\text{x}}{\text{x}^3}$
$=4\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)^3$
$=4\times1$
$=4$
$=\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}\big(3\sin\text{x}-4\sin^3\text{x}\big)}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{4\sin^3\text{x}}{\text{x}^3}$
$=4\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\Big)^3$
$=4\times1$
$=4$
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