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If \(\mathop {\lim }\limits_{x \to \infty } \left( {ax + \left( {\frac{{7 - \sqrt 3 {x^2}}}{{3 - x}}} \right)} \right) = b,\) a finite number, then the values of a and b are:
1. a = -√3, b = 3√3
2. a = 3, b = √3
3. a = 2√3, b = -3
4. a = -2√3, b = 2√3

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Correct Answer - Option 1 : a = -√3, b = 3√3

\(\mathop {\lim }\limits_{x \to \infty } \left( {ax + \left( {\frac{{7 - \sqrt 3 {x^2}}}{{3 - x}}} \right)} \right) = b\;\)

\(ax + \frac{{7 - \sqrt 3 {x^2}}}{{3 - x}}\)

\( = \frac{{ax\left( {3 - x} \right) + 7 - \sqrt 3 \;{x^2}}}{{3 - x}}\)

\( = \frac{{3ax - a{x^2} + 7 - \sqrt 3 \;{x^2}}}{{3 - x}}\)

\( = \frac{{ - \sqrt 3 \;{x^2} - a{x^2} + 3ax + 7}}{{3 - x}}\)

\(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\left( { - \sqrt 3 - a} \right) + x\left( {3a + \frac{7}{x}} \right)}}{{x\left( {\frac{3}{x} - 1} \right)}}\)

∴ For the limit to be finite, the x2 term in the numerator should be 0.
\(\therefore \; - \sqrt 3 - a = 0\)

\(a = - \sqrt 3 \)

Now,

\(\mathop {\lim }\limits_{x \to \infty } \frac{{x\left( { - 3\sqrt 3 + \frac{7}{x}} \right)}}{{x\left( {\frac{3}{x} - 1} \right)}}\)

\( = \frac{{ - 3\sqrt 3 }}{{ - 1}} = 3\sqrt 3 \)

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