Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(2+\text{x})-\sin(2-\text{x})}{\text{x}}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(2+\text{x})-\sin(2-\text{x})}{\text{x}}$
✓
Answer
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(2+\text{x})-\sin(2-\text{x})}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}2\cos\frac{\Big(\frac{{2+\text{x}+2-\text{x}}{2}}{\text{x}}\Big)\times\sin\Big(\frac{2+\text{x}-2+\text{x}}{2}\Big)}{{\text{x}}}$
$=2\lim\limits_{\text{x}\rightarrow0}\frac{\cos(2)\times\sin\text{x}}{\text{x}}$
$=2\lim\limits_{\text{x}\rightarrow0}\cos2\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=2\cos2\times1$
$=2\cos2$
$=\lim\limits_{\text{x}\rightarrow0}2\cos\frac{\Big(\frac{{2+\text{x}+2-\text{x}}{2}}{\text{x}}\Big)\times\sin\Big(\frac{2+\text{x}-2+\text{x}}{2}\Big)}{{\text{x}}}$
$=2\lim\limits_{\text{x}\rightarrow0}\frac{\cos(2)\times\sin\text{x}}{\text{x}}$
$=2\lim\limits_{\text{x}\rightarrow0}\cos2\times\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=2\cos2\times1$
$=2\cos2$
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