Evaluate the following limit: _ x 2 x^2-x-2 x^2-2x+ (x-2) - Vidyadip

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^2-\text{x}-2}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$

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Answer

$\lim\limits_{\text{x}\rightarrow2}\frac{\text{x}^2-\text{x}-2}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{(\text{x}-2)(\text{x}+1)}{\text{x}^2-2\text{x}+\sin(\text{x}-2)}$
$=\lim\limits_{\text{x}\rightarrow2}\frac{1}{\frac{\text{x}}{\text{x}+1}+\frac{\sin(\text{x}-2)}{(\text{x}-2)(\text{x}+1)}}$
$=\lim\limits_{\text{x}\rightarrow2}(\text{x}+1)\Bigg(\frac{1}{\text{x}+\frac{\sin(\text{x}-2)}{\text{x}-2}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow2}(\text{x}+1)\frac{1}{\lim\limits_{\text{x}\rightarrow2}(\text{x})+\lim\limits_{\text{x}\rightarrow2}\frac{\sin(\text{x}-2)}{\text{x}-2}}$
$=(2+1)\times\frac{1}{(2)\lim\limits_{\text{x}\rightarrow2\rightarrow0}\frac{\sin(\text{x}-2)}{(\text{x}-2)}}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=3\times\frac{1}{2+1}$
$=1$

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