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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$
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Answer

$\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$$=\lim\limits_{\text{x}\rightarrow2}\Bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}\big(\text{x}^2-2\text{x}-\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}-\frac{2(2\text{x}-3)}{\text{x}(\text{x}-2)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}(\text{x}-1)-2(2\text{x}-3)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}^2-\text{x}-4\text{x}+6}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}^2-2\text{x}-3\text{x}+6}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{\text{x}(\text{x}-2)-3(\text{x}-2)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{(\text{x}-2)(\text{x}-3)}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow2}\bigg\{\frac{1}{\text{x}-2}\frac{2(2\text{x}-3)}{\text{x}(\text{x}-2)(\text{x}-1)}\bigg\}$
$=\frac{-1}{2}$

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