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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$
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Answer

$\lim\limits_{\text{x}\rightarrow-1}\frac{\text{x}^2-\text{x}-2}{\big(\text{x}^2+\text{x}\big)+\sin(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{(\text{x}-2)(\text{x}+2)}{\text{x}(\text{x}+1)+\sin(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\frac{\text{x}}{\text{x}-2}+\frac{\sin(\text{x}+1)}{(\text{x}-2)(\text{x}+1)}}$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{(\text{x}-2)}\Bigg(\frac{1}{\text{x}+\frac{\sin(\text{x}+1)}{\text{x}+1}}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow-1}\frac{1}{\text{x}-2}\times\frac{1}{\lim\limits_{\text{x}\rightarrow-1}(\text{x})+\lim\limits_{\text{x}\rightarrow-0}\sin\frac{\text{x}+1}{\text{x}+1}}$
$=\Big(\frac{1}{-1-2}\Big)\times\frac{1}{(-1)+1}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac{1}{0}$ $\Big[\because\frac10=\infty\Big]$
$=\infty$

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