Math, asked by Anonymous, 2 days ago

\bf\red{QƲƐŞƬiøƝ}
\bf\purple{\bf{\displaystyle\lim_{x \to 1}\rm} \: \: \: \frac{ {x}^{4} - 3 {x}^{2} + 2 }{ {x}^{3} - 5 {x}^{2} + 3x + 1} }

{\boxed{\underline\mathfrak\green{Don't \: \: Spam♡︎}}}
{\boxed{\underline{\sf\pink{Need \: \: quality \: \: answer♡︎}}}}

Answers

Answered by Anonymous
42

{\underline{\boxed{\mathfrak\red{QƲƐŞƬiøƝ:-}}}}

\bf\purple{\bf{\displaystyle\lim_{x \to 1}\rm} \: \: \: \frac{ {x}^{4} - 3 {x}^{2} + 2 }{ {x}^{3} - 5 {x}^{2} + 3x + 1} }

{\underline{\boxed{\mathfrak\pink{Answer:-}}}}

\bf\red{Let \: \: {x}^{2} = \: t}

↦ \: \bf{ {t}^{2} - 3t + 2}

↦ \: \bf{ {t}^{2} - 2t - t+ 2 }

↦ \: \bf{t(t - 2) - 1(t - 2)}

↦ \: \bf{(t - 2)(t - 1)}

↦ \: \bf{ ({x}^{2} - 2)( {x}^{2} - 1) }

↦\bf{ ({x}^{2} - 2)(x + 1)(x - 1)}

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

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\bf{ {x}^{3} - 5 {x}^{2} + 3x + 1}

↦\bf{ {x}^{2} (x - 1) - 4x(x - 1) - 1(x - 1)}

↦ \: \bf{(x - 1)( {x}^{2} - 4x - 1) }

↦ \: \bf{(x - 1)[x - (2 + \sqrt{5}] [x - (2 - \sqrt{5)} ]}

N.T.L

ㅤㅤㅤㅤㅤNow taking limit,

⇒ \: \bf{ \frac{ - 1 \times 2}{(1 - 2 - \sqrt{5} )(1 - 2 + \sqrt{5}) }}

⇒ \: \bf{ \frac{ - 2}{( - 1 - \sqrt{5})( - 1 + \sqrt{5} ) } }

⇒ \: \bf{ \frac{2}{( \sqrt{5} + 1)( \sqrt{5} - 1) } }

⇒ \: \bf{ \frac{2}{5 - 1} }

⇒ \: \bf{ \frac{2}{4} }

⇒ \: \pink{\boxed{ \frac{1}{2} }}

Answered by IIMrSAURAVII
3

Answer:

Letx2=t

↦ \: \bf{ {t}^{2} - 3t + 2}↦t2−3t+2

↦ \: \bf{ {t}^{2} - 2t - t+ 2 }↦t2−2t−t+2

↦ \: \bf{t(t - 2) - 1(t - 2)}↦t(t−2)−1(t−2)

↦ \: \bf{(t - 2)(t - 1)}↦(t−2)(t−1)

↦ \: \bf{ ({x}^{2} - 2)( {x}^{2} - 1) }↦(x2−2)(x2−1)

↦\bf{ ({x}^{2} - 2)(x + 1)(x - 1)}↦(x2−2)(x+1)(x−1)

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

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\bf{ {x}^{3} - 5 {x}^{2} + 3x + 1}x3−5x2+3x+1

↦\bf{ {x}^{2} (x - 1) - 4x(x - 1) - 1(x - 1)}↦x2(x−1)−4x(x−1)−1(x−1)

↦ \: \bf{(x - 1)( {x}^{2} - 4x - 1) }↦(x−1)(x2−4x−1)

↦ \: \bf{(x - 1)[x - (2 + \sqrt{5}] [x - (2 - \sqrt{5)} ]}↦(x−1)[x−(2+5][x−(2−5)]

N.T.L

ㅤㅤㅤㅤㅤNow taking limit,

⇒ \: \bf{ \frac{ - 1 \times 2}{(1 - 2 - \sqrt{5} )(1 - 2 + \sqrt{5}) }}⇒(1−2−5)(1−2+5)−1×2

⇒ \: \bf{ \frac{ - 2}{( - 1 - \sqrt{5})( - 1 + \sqrt{5} ) } }⇒(−1−5)(−1+5)−2

⇒ \: \bf{ \frac{2}{( \sqrt{5} + 1)( \sqrt{5} - 1) } }⇒(5+1)(5−1)2

⇒ \: \bf{ \frac{2}{5 - 1} }⇒5−12

⇒ \: \bf{ \frac{2}{4} }⇒42

⇒ \: \pink{\boxed{ \frac{1}{2} }}⇒21

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