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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-4\text{x}+3}{{\text{x}^2}-2\text{x}-3}$
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Answer

$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-4\text{x}+3}{{\text{x}^2}-2\text{x}-3}$$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^2-3\text{x}-\text{x}+3}{{\text{x}^2}+\text{x}-3\text{x}-3}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}(\text{x}-1)-3(\text{x}-1)}{\text{x}({\text{x}}+1)-3(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-1)(\text{x}-3)}{({\text{x}}+1)(\text{x}-3)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-1}{\text{x}+1}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}(\text{x}-1)-3(\text{x}-1)}{\text{x}({\text{x}}+1)-3(\text{x}+1)}$
$=\frac{3-1}{3+1}$
$=\frac{2}{4}=\frac12$

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