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प्रश्न
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
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उत्तर
Let p(x) = x3 + 3x2 \[-\] 6x + 2
p(1) = 1 + 3\[-\] 6 + 2
= 0
Now,
\[\left( x + 2 \right)\] is a factor of p(x).
p(x) = (x\[-\] 1)(x2 + 4x \[-\] 2)
q(x) = x3 + 3x2\[-\]3x + 2
q(1) = 1 + 3 \[-\]3 \[-\]1
= 0
\[Now, \left( x + 2 \right)\] is a factor of p(x).
\[\Rightarrow \lim_{x \to 1} \left[ \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( x^2 + 4x - 2 \right)}{\left( x - 1 \right)\left( x^2 + 4x + 1 \right)} \right]\]
\[ = \frac{(1 )^2 + 4 \times 1 - 2}{\left( 1 \right)^2 + 4 \times 1 + 1}\]
\[ = \frac{1 + 4 - 2}{1 + 4 + 1}\]
\[ = \frac{3}{6}\]
\[ = \frac{1}{2}\]
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