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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}-2}{\text{x}^2-\text{x}}-\frac{1}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$
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Answer

$\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}-2}{\text{x}^2-\text{x}}-\frac{1}{\text{x}^2-3\text{x}^2+2\text{x}}\bigg\}$$=\lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1)}{\text{x}\big(\text{x}^2-3\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}\big(\text{x}^2-3\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}\big(\text{x}^2-1\text{x}-2\text{x}+2\big)}\Bigg\}$
$= \lim\limits_{\text{x}\rightarrow1}\Bigg\{\frac{\text{x}-2}{\text{x}(\text{x}-1)}-\frac{1}{\text{x}(\text{x}-1)(\text{x}-2)}\Bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{(\text{x}-2)^2-1}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2+4-4\text{x}-1}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2-4\text{x}+3}{\text{x}(\text{x}-1)(\text{x}-2)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\bigg\{\frac{\text{x}^2-4\text{x}+3}{\text{x}(\text{x}-2)(\text{x}-1)}\bigg\}$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{\text{x}^2-\text{x}-3\text{x}+3}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{\text{x}(\text{x}-1)-3(\text{x}-1)}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\lim\limits_{\text{x}\rightarrow1}\Big[\frac{(\text{x}-3)(\text{x}-1)}{\text{x}(\text{x}-1)(\text{x}-2)}\Big]$
$=\frac{(1-3)}{1(1-2)}$
$=\frac{-2}{-1}$
$=2$

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