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प्रश्न
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
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उत्तर
\[\lim_{x \to 2} \left[ \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6} \right]\] It is of the form \[\frac{0}{0} .\]
Let p(x) = x3 + 3x2 \[-\]9x\[-\] 2
p(2) = 8 + 12 \[-\]18 \[-\]2
= 0
Now,
\[\left( x - 2 \right)\] is a factor of p(x).
Let q(x) = x3 – x – 6
q(2) = 8 – 2 – 6
= 0
\[Now, \left( x - 2 \right)\]is a factor of q(x).
\[\Rightarrow \lim_{x \to 2} \left[ \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x^2 + 5x + 1 \right)}{\left( x - 2 \right)\left( x^2 + 2x + 3 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 + 5x + 1}{x^2 + 2x + 3} \right]\]
\[ = \frac{(2 )^2 + 5 \times 2 + 1}{\left( 2 \right)^2 + 2 \times 2 + 3}\]
\[ = \frac{4 + 10 + 1}{4 + 4 + 3}\]
\[ = \frac{15}{11}\]
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