Question

the value of lim x to 2 x^5 -32/x^2-4​

Answer 1

$\bf{Given}$

$\tt \to \displaystyle \lim _{ \tt \: x \to2} \tt\frac{ {x}^{5} - 32}{ {x}^{2} - 4}$

$\bf \: Now \: check \: the \: form \: of \: limit \: so \: put \: x = 2$

$\tt \to \dfrac{(2) {}^{5} - 32}{(2)^{2} - 4} = \dfrac{0}{0}$

$\bf \: So \: it \: is \: \dfrac{0}{0} \: form \: now \: we \: use \: L'hospital \: rule$

$\bf \: Differentiate \: W.R.T \: x \: on \: both \: numerator \: and \: denominator </p><p>$

$\tt \to \: \displaystyle \: \lim_{ \tt \: x \to 2} \tt\frac{ \dfrac{d( {x}^{5} - 32) }{dx} }{ \dfrac{d( {x}^{2} - 4)}{dx} }$

$\tt \to \displaystyle \lim _{ \tt \: x \to2} \tt\frac{5 {x}^{4} }{2x}$

$\bf \: Now \: put \: x = 2$

$\tt \to \dfrac{5 {x}^{3} }{2} = \dfrac{5(2)^{3} }{2}$

$\tt \to \: 5(2) {}^{2} = 5 \times 4 = 20$

$\bf \: Answer$

$\tt = 20$

$\bf \: \underline{Some \:Formula \: of \: Differentiation }</p><p>$

$\tt \to \: \dfrac{d {x}^{n}}{dx} = nx ^{n - 1}$

$\tt \to \: \dfrac{d(2)}{dx} = 0$

$\tt \to \: \dfrac{d( {e}^{x}) }{dx} = {e}^{x}$

$\tt \to \: \dfrac{d(lnx)}{dx} = \dfrac{1}{x}$

$\tt \to \dfrac{d(sinx)}{dx} = cosx$

$\tt \to \: \dfrac{d(cosx)}{dx} = - sinx$

$\tt \to \: \dfrac{d(tanx)}{dx} = sec {}^{2} x$

Answer 2

TO SOLVE.

• $\\ \\ : \implies \displaystyle \lim _ { \: \: \tt \: x \to2} \tt(\frac{ {x}^{5} - 32}{ {x}^{2} - 4})$

EXPLANATION.

$\\\\ \frak{ \underbrace{ \underline{ \: \: \:Firstly ,\: \: \: put \: \: \: \boxed{ \bold x}, \: \: \: }}}$

• $\ \\ \tt : \implies \dfrac{(2) {}^{5} - 32}{(2)^{2} - 4} = \dfrac{0}{0}$

$\\ \\ \frak{ \underbrace{ \underline{ \: Secondly,\: use \: \bf{ \: L'hospital \: rule, }}}}$

$\frak{ \underbrace{ \underline{ Thirdly,\: differentiate \: \bf \: W.R.T \: x \: \frak{on \: both \: numerator \: and \: denominator,}}}}$

$\sf : \implies \: \displaystyle \: \lim_{ \tt \: x \to 2} \: \sf(\frac{ \dfrac{d( {x}^{5} - 32) }{dx} }{ \dfrac{d( {x}^{2} - 4)}{dx} })$

$\sf : \implies \displaystyle \lim _{ \: \tt \: x \to2} \tt(\frac{5 {x}^{4} }{2x})$

$\frak{ \underbrace{ \underline{ \: Fourthly, \: put \: \boxed{ \bf \: x = 2}}}}$

$\sf : \implies \dfrac{5 {x}^{3} }{2} = \dfrac{5(2)^{3} }{2}$

$\underline{\boxed{\frak{: \implies \: 5(2) {}^{2} = 5 \times 4 = 20}}}$

$\underline{\overbrace{ \bf \therefore, \: \sf the \: value \: of \: \longmapsto \displaystyle \lim _ { \: \: \tt \: x \to2} \tt(\frac{ {x}^{5} - 32}{ {x}^{2} - 4}) \: \: \sf is \: \bf \: 20.}}$

✬ Hope it helps u ✬