Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$

Answer

 $\lim\limits_{\text{x}\rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}$

$=\lim\limits_{\text{x} \rightarrow0}\frac{3\sin^2\text{x}-2\sin\text{x}^2}{3\text{x}^2}-\lim\limits_{\text{x} \rightarrow0}\frac{2\sin\text{x}^2}{3\text{x}^2}$

$=\lim\limits_{\text{x} \rightarrow0}\big(\frac{\sin\text{x}}{\text{x}}\big)^2-\frac{2}{3}\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}^2}{\text{x}^2}$

$=1-\frac{2}{3}\times1$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$

$=1-\frac23$

$=\frac{3-2}{3}$

$=\frac{1}{3}$ 

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