Question

find
$\displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 }$
​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 }=\frac{3}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } \\ \\ \tt \circ \: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab) \\ \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ (x - 1)( {x}^{2} + {1}^{2} + x) }{ (x - 1)({x}^{2} + x + 2)} \\ \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ ({x}^{2} + x + 1) }{ ({x}^{2} + x + 2) } \\ \\ \tt: \implies \frac{\displaystyle \lim_{x \to 1} \: \: ( {x}^{2} + x + 1) }{\displaystyle \lim_{x \to 1} \: \: ( {x}^{2} + x + 2) } \\ \\ \tt: \implies \frac{ {1}^{2} + 1 + 1}{ {1}^{2} + 1 + 2} \\ \\ \green{ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } = \frac{3}{4} }$

Answer 2

Answer:

$</p><p>\begin{gathered}\pink{\tt{\therefore{\displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 }=\frac{3}{4}}}}\\\end{gathered}$

∴

$\blue{\huge{\underline{\underline{Step-by-step\: \: explanation:}}}}$

$\begin{gathered}\pink{\underline \bold{Given :}} \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } \\ \\ \orange{\underline \bold{To \: Find :}} \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } = ?\end{gathered} </p><p>$

• According to given question :

$\begin{gathered}\bold{As \: we \: know \: that} \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } \\ \\ \tt \circ \: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab) \\ \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ (x - 1)( {x}^{2} + {1}^{2} + x) }{ (x - 1)({x}^{2} + x + 2)} \\ \\ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ ({x}^{2} + x + 1) }{ ({x}^{2} + x + 2) } \\ \\ \tt: \implies \frac{\displaystyle \lim_{x \to 1} \: \: ( {x}^{2} + x + 1) }{\displaystyle \lim_{x \to 1} \: \: ( {x}^{2} + x + 2) } \\ \\ \tt: \implies \frac{ {1}^{2} + 1 + 1}{ {1}^{2} + 1 + 2} \\ \\ \blue{ \tt: \implies \displaystyle \lim_{x \to 1} \: \: \frac{ {x}^{3} - 1 }{ {x}^{3} + x - 2 } = \frac{3}{4} }\end{gathered}$

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