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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$
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Answer

$\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\text{x}^2+3\text{x}-6}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+\text{x}-2\big]}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3\big[\text{x}^2+2\text{x}-\text{x}-2\big]}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)(\text{x}-1)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)\big(\sqrt{\text{x}}-1\big)}{3(\text{x}+2)\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{(2\text{x}-3)}{3(\text{x}+2)\big(\sqrt{\text{x}}+1\big)}$ $=\frac{(2-3)}{3(1+2)(1+1)}$ $=\frac{-1}{3\times3\times2}$ $=-\frac{1}{18}$

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