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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits2 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3\text{x}^2-4\text{x}^2+6\text{x}-1}{2\text{x}^3+\text{x}^2-5\text{x}+7}$
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Answer

$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3\text{x}^2-4\text{x}^2+6\text{x}-1}{2\text{x}^3+\text{x}^2-5\text{x}+7}$ $=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3-\frac{4}{\text{x}}+\frac{6}{\text{x}^2}-\frac{1}{\text{x}^3}}{2+\frac{1}{\text{x}}-\frac{5}{\text{x}^2}+\frac{7}{\text{x}^3}}$ $=\frac{3-0+0-0}{2+0-0+0}$ $=\frac32$

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