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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits2 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}^2+\text{x}-2}-\frac{\text{x}}{\text{x}^3-1}\Big)$
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Answer

$\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}^2+\text{x}-2}-\frac{\text{x}}{\text{x}^3-1}\Big)$$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{1}{\text{x}^2+\text{x}-2}-\frac{\text{x}}{(\text{x}-1)(\text{x}^2+\text{x}+1)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}^3-1-\text{x}^3-\text{x}^2+2\text{x}}{(\text{x}^3-1)(\text{x}^2+\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Bigg(\frac{\big(\text{x}^2-2\text{x}+1\big)}{\big(\text{x}^3-1\big)\big(\text{x}^2+\text{x}-2\big)}\Bigg)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{(\text{x}-1)(\text{x}-1)}{(\text{x}-1)(\text{x}^2+1+\text{x})(\text{x}^2+\text{x}-2)}\Big)$
$=\lim\limits_{\text{x}\rightarrow1}\Big(\frac{\text{x}-1}{(\text{x}^2+1+\text{x})(\text{x}+2)(\text{x}-1)}\Big)$
$=\frac{1}{(1+1+1)(1+2)}$
$=\frac19$

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