Question

$\displaystyle \lim_{x \to0} \frac{ \int \limits_{0}^{{x}^{2}} \sqrt{4 + {t}^{3} } dt }{ {x}^{2} }$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle \lim_{x \to0} \frac{ \displaystyle\int \limits_{0}^{{x}^{2}} \sqrt{4 + {t}^{3} } dt }{ {x}^{2} }$

On applying L Hospital Rule, we get

$\rm \: = \:\displaystyle \lim_{x \to0} \frac{\dfrac{d}{dx} \displaystyle\int \limits_{0}^{{x}^{2}} \sqrt{4 + {t}^{3} } dt }{\dfrac{d}{dx} {x}^{2} }$

Now, we know,

$\rm \: \dfrac{d}{dx} {x}^{2} = 2x \:$

We know, Differentiation under the integral sign, using Leibnitz Rule

$\rm \: \dfrac{d}{dx}\displaystyle\int \limits_{a}^{b}f(x,y)dy = \displaystyle\int \limits_{a}^{b}\dfrac{ \partial \: }{\partial\: x}f(x,y)dy + \dfrac{db}{dx}f(x,b) - \dfrac{da}{dx}f(x,a)$

So, using this rule,

$\rm \: \dfrac{d}{dx}\displaystyle\int \limits_{0}^{{x}^{2}} \sqrt{4 + {t}^{3} } dt$

$\rm \: = \:\displaystyle\int \limits_{0}^{{x}^{2}} \dfrac{\partial}{\partial \: x} \: \sqrt{4 + {t}^{3} } dt + \dfrac{d}{dx}( {x}^{2}) \times \sqrt{4 + {x}^{3} } - \dfrac{d}{dx}(0) \sqrt{4 + {0}^{3} }$

$\rm \: = \:0 + 2x \times \sqrt{4 + {x}^{3} } - 0$

$\rm \: = \: 2x\sqrt{4 + {x}^{3} }$

So, on substituting all these values in above expression, we get

$\rm \: = \:\displaystyle\lim_{x \to \: 0} \: \frac{2x \sqrt{4 + {x}^{3} } }{2x}$

$\rm \: = \:\displaystyle\lim_{x \to \: 0} \: \sqrt{4 + {x}^{3} }$

$\rm \: = \: \sqrt{4}$

$\rm \: = \:2$

Hence,

$\rm\implies \: \boxed{\sf{  \:\: \rm \: \displaystyle \lim_{x \to0} \frac{ \displaystyle\int \limits_{0}^{{x}^{2}} \sqrt{4 + {t}^{3} } dt }{ {x}^{2} } = 2 \: \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\huge\colorbox{white}{ \colorbox{white}{\colorbox{white}{Question}}}$

$\Huge\sf\implies$ $\displaystyle \lim_{x \to0} \frac{ \int \limits_{0}^{{x}^{2}} \sqrt{4 + {t}^{3} } dt }{ {x}^{2} }$

_______________________________

$\huge\sf\star$ Solution :-

$\Huge\sf\implies$ $\displaystyle\color{gray} \lim_{x \to0} \frac{ \int \limits_{0}^{{x}^{2}} \sqrt{4 + {t}^{3} } dt }{ {x}^{2} }$

$\Huge\sf\implies$ If we put x = 0 , then $\frac{0}{0}$ form

So, by L.Hospital rule,

$\lim_{x\rightarrow 0} [tex]\sqrt{4\:+\:6²)³}$

$\Huge\sf\implies$ $\frac{2}{2n}$ (2²) - 0....

$\Huge\sf\implies$$\bold\color{red} {Using\: Newton\:Leibnitz\:Formula\:in\:numerator}$

$\huge\lim_{x\rightarrow0} [tex] \sqrt\frac{4\:+\:x⁶ \:. 2x }{2x}}$

$\Huge\sf\implies$ $\lim_{x\rightarrow 0}$ $\sqrt{4\:+\:x²}$

$\Huge\sf\implies$ $\sqrt{4\: + \:0}$

$\sf\huge\implies$ 2

Hence , the value of the limit is 2

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