Question

Find the value of
$\displaystyle\bf\:\lim_{\Delta x \to \: 0} \: \frac{3(x + \Delta x)^{2} - 3x^{2}}{\Delta x}$
​

Answer 1

GIVEN :-

$\\ \red \bigstar \displaystyle \bf \: \lim_{ \ \Delta x \to \: 0} \: \dfrac{3(x +\Delta x) ^{2} - 3x ^{2} }{\Delta \: x}$

TO FIND :-

$\\ \red \bigstar \: \displaystyle \bf \: value \: of \lim_{ \ \Delta x \to \: 0} \: \dfrac{3(x +\Delta x) ^{2} - 3x ^{2} }{\Delta \: x}$

SOLUTION :-

$\\ : \implies\displaystyle \bf \: \lim_{ \ \Delta x \to \: 0} \: \dfrac{3(x +\Delta x) ^{2} - 3x ^{2} }{\Delta \: x}$

$\orange \bigstar \blue{ \tt \:By \: using \: identity :- (a + b)^{2} = a^{2} + 2ab + b^{2} }$

$\displaystyle \bf \: : \implies \lim_{ \ \Delta x \to \: 0} \: \dfrac{3 \{x^{2} +2 \times x \times \Delta x) + (\Delta x)^{2} \} - 3x ^{2} }{\Delta x}$

$\displaystyle \bf : \implies\: \lim_{ \ \Delta x \to \: 0} \: \dfrac{3(x^{2} +2 x \Delta x+ \Delta x .\Delta x) - 3x ^{2} }{\Delta x}$

$\displaystyle \bf : \implies\: \lim_{ \ \Delta x \to \: 0} \: \dfrac{3x ^{2} + 6x\Delta x + 3\Delta x\Delta x - 3x^{2} }{\Delta x}$

$\displaystyle \bf : \implies\: \lim_{ \ \Delta x \to \: 0} \: \dfrac{3x ^{2} - 3x ^{2} + 6x\Delta x + 3\Delta x\Delta x }{\Delta x}$

$\displaystyle \bf : \implies\: \lim_{ \ \Delta x \to \: 0} \: \dfrac{ 6x \cancel{\Delta x } + 3\Delta x \cancel{\Delta x } }{ \cancel{\Delta x}}$

$\displaystyle \bf : \implies\: \lim_{ \ \Delta x \to \: 0} \: 6x + 3\Delta x$

$\orange \bigstar \blue{ \tt \: On \: Removing \: Limit.}$

$\displaystyle \bf : \implies\: 6x + 0$

$\red \bigstar \: \underline{\boxed{ \displaystyle \bf \: \lim_{ \ \Delta x \to \: 0} \: \dfrac{3(x +\Delta x) ^{2} - 3x ^{2} }{\Delta \: x} = 6x}}$

Answer 2

$\red{\bold{\underline{\underline{Answer:}}}} </p><p></p><p>$

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

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} </p><p>$

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

• According to given question :

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

☃️Hence We Are done !!

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