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प्रश्न
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
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उत्तर
p(x) = x4\[-\] 3x3 + 2
p(1) = 1\[-\]3 + 2
= 0
Now,
\[\left( x - 1 \right)\] is a factor of p(x).
q(x) = x3 \[-\] 5x2 + 3x + 1
q(1) = 1\[-\]5 + 3 + 1
= 0
\[\text{ Now }, \left( x - 1 \right)\]is a factor of q(x).
\[\Rightarrow \lim_{x \to 1} \left[ \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( x^3 - 2 x^2 - 2x - 2 \right)}{\left( x - 1 \right)\left( x^2 - 4x - 1 \right)} \right]\]
\[ = \frac{(1 )^3 - 2 \left( 1 \right)^2 - 2\left( 1 \right) - 2}{\left( 1 \right)^2 - 4 \times 1 - 1}\]
\[ = \frac{1 - 2 - 2 - 2}{1 - 4 - 1}\]
\[ = \frac{- 5}{- 4}\]
\[ = \frac{5}{4}\]
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