Question

$\color{blue}{\rm{ \displaystyle \lim _{ \rm \: x \to1} \frac{ \displaystyle\int_{0} ^{( \rm \: x - 1) ^{2} } \rm t \cos tdt}{ \rm(x - 1) \sin(x - 1)} } }$​

Answer 1

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

Given expression is

$\rm \: \displaystyle \lim _{ \rm \: x \to1} \frac{ \displaystyle\int_{0} ^{( \rm \: x - 1) ^{2} } \rm t \cos tdt}{ \rm(x - 1) \sin(x - 1)}$

If we substitute directly x = 1, we get indeterminant form.

So, above expression can be rewritten as

$\rm \: = \: \displaystyle \lim _{ \rm \: x \to1} \frac{ \displaystyle\int_{0} ^{( \rm \: x - 1) ^{2} } \rm t \cos tdt}{ \rm(x - 1) ^{2} \times \dfrac{sin(x - 1)}{(x - 1)} }$

We know,

$\boxed{ \rm{ \:\displaystyle \lim _{ \rm \: x \to0}\rm \: \frac{sinx}{x} \: = \: 1 \: }}$

So, using this identity, we get

$\rm \: = \: \displaystyle \lim _{ \rm \: x \to1} \frac{ \displaystyle\int_{0} ^{( \rm \: x - 1) ^{2} } \rm t \cos tdt}{ \rm(x - 1) ^{2}}$

On applying L Hospital Rule, we get

$\rm \: = \: \displaystyle \lim _{ \rm \: x \to1} \frac{ \dfrac{d}{dx} \displaystyle\int_{0} ^{( \rm \: x - 1) ^{2} } \rm t \cos tdt}{\dfrac{d}{dx} \rm(x - 1) ^{2}}$

So, on differentiating numerator using Leibnitz rule, we get

$\rm \: = \: \displaystyle \lim _{ \rm \: x \to1} \frac{ \rm {(x - 1)}^{2} \cos {(x - 1)}^{2}\dfrac{d}{dx} {(x - 1)}^{2} }{2(x - 1)}$

$\rm \: = \: \displaystyle \lim _{ \rm \: x \to1} \frac{ \rm {(x - 1)} [\cos {(x - 1)}^{2}] \: \times \: 2(x - 1)}{2}$

$\rm \: = \: \displaystyle \lim _{ \rm \: x \to1} \: \rm \: {(x - 1)}^{2} \: cos {(x - 1)}^{2}$

$\rm \: = \: {(1 - 1)}^{2} \: cos {(1 - 1)}^{2}$

$\rm \: = \: 0 \times cos0$

$\rm \: = \: 0 \times 1$

$\rm \: = \: 0$

Hence,

$\color{green}\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle \lim _{ \rm \: x \to1} \frac{ \displaystyle\int_{0} ^{( \rm \: x - 1) ^{2} } \rm t \cos tdt}{ \rm(x - 1) \sin(x - 1)} = 0 \: \: }}$

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Formulae Used :-

Leibnitz Rule :-

$\rm \: \dfrac{d}{dx}\displaystyle\int_{a} ^{\rm b}f(x,t)dt = \displaystyle\int_{a} ^{\rm b} \: \frac{\partial }{\partial x} f(x,t)dt + \dfrac{db}{dx}[f(x,b)] - \dfrac{da}{dx}[f(x,a)]$

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

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Additional Information :-

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