Math, asked by piyushsharma82paxg79, 3 months ago

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Answered by MagicalBeast
12

Solution :

 \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{x \sqrt{x} \:   -   \: a\sqrt{a} }{x - a}  \\  \\  \sf \: if \: we \: put \: x = a \: we \: get \\  \sf \:  \implies \:  \dfrac{a \sqrt{a} - a \sqrt{a}  }{a - a}  \:  =  \:  \dfrac{0}{0}

Therefore applying , L-Hospital rule

\sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{x \sqrt{x} \:   -   \: a\sqrt{a} }{x - a}  \\ \\   \sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{ \dfrac{d}{dx}  \bigg(x \sqrt{x} \:   -   \: a\sqrt{a} \bigg) }{ \dfrac{d}{dx} (x - a)}  \\  \\   \sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{ \dfrac{d}{dx}  \bigg(x \sqrt{x} \:   \bigg) -   \:  \dfrac{d}{dx}  \bigg(a\sqrt{a} \bigg) }{ \dfrac{d}{dx} (x) -  \dfrac{d}{dx} (a)}

Note - Here a is constant, and diffrenciation of constant w.r.t x is zero

 \sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{ \dfrac{d}{dx}  \bigg(x \sqrt{x} \:   \bigg) - 0}{ \dfrac{d}{dx} (x) -  0} \\  \\  \sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{ \dfrac{d}{dx}  \bigg(x \sqrt{x} \:   \bigg) }{ \dfrac{d}{dx} (x) }

Note - Use products rule of diffrenciation in numerator

\sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{  \bigg( \sqrt{x} \times  \dfrac{d}{dx}  (x) \bigg)  \: +   \:  \bigg(x \:  \times \:  \dfrac{d}{dx}( \sqrt{x} )  \bigg)  }{ \dfrac{d}{dx} (x) }

Note - identity used

\sf \bullet  \dfrac{d}{dx}  {x}^{m}  \:  = m  \times {x}^{(m - 1)}

\sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \:  \sf \dfrac{  \bigg( \sqrt{x} \times  1 \bigg)  \: +   \:  \bigg(x \:  \times \:   \dfrac{1}{2 \sqrt{x} }  \bigg)  }{ 1 } \\  \\  \sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \: \bigg(  \sf   \sqrt{x}  \: +   \:   \:   \dfrac{x}{2 \sqrt{x} }  \bigg)   \\  \\ \sf \implies \: \sf \displaystyle \:  \lim_{ \sf \: x \to \: a} \: \bigg(  \sf   \sqrt{x}  \: +   \:   \:   \dfrac{ \sqrt{x} }{2  }  \bigg)

Now in above put limit { that is x = a }

 \sf \implies \:  \sqrt{a}  +   \dfrac{ \sqrt{a} }{2}  \\  \\  \sf \implies \:  \sqrt{a}  \bigg(1 +  \dfrac{1}{2}  \bigg) \\  \\  \sf \implies \:  \sqrt{a}  \bigg( \dfrac{3}{2}  \bigg) \\  \\  \sf \implies \:   \bold{\dfrac{3 \sqrt{a} }{2} }

ANSWER : (3/2)√a


Anonymous: Outstanding! :)
piyushsharma82paxg79: thanks
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