Question

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Answer 1

Question :

Find the value of :

$\rm\lim{x\to1}\:\dfrac{\int\limits_{0}^{(x-1)^2}t\cos(t)dt}{(x-1)\sin(x-1)}$

Formula's used

1)Leibnitz's Rule : Derivative of Definate integral

Let g be a continuous function on [a,b] and α(x) ,ß(x) are differentiable functions of x whose value lie in [a,b] , then

$\sf\dfrac{d}{dx}\int\limits_{\alpha(x)}^{\beta(x)}g(t)=g[\beta(x)]\beta'(x)-g[\alpha(x)]\alpha'(x)$

2) Chain rule:

Let y=f(t) and t=g(x) , then

$\sf\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{dx}$

3) L'hospital rule :

When a function is in 0/0 (or) ∞/∞ form then both the numerator and denominator of the function can be differentiated to evaluate the limit .

Solution :

We have;

$\sf\lim{x\to1}\:\dfrac{\int\limits_{0}^{(x-1)^2}t\cos(t)dt}{(x-1)\sin(x-1)}$

Clearly it is $\sf\dfrac{0}{0}$ form

$\sf\blue{So\:,we\:can\:solve\:it\:by\:l-hosptial\:rule}$

Now , let's solve the problem

$\sf\lim{x\to1}\:\dfrac{\int\limits_{0}^{(x-1)^2}t\cos(t)dt}{(x-1)\sin(x-1)}$

by l-hosptial rule :

$\sf\lim{x\to1}\:\dfrac{\frac{d(x-1)^2}{dx}\times(x-1)^2\cos(x-1)^2-0}{(x-1)\times\frac{d(\sin(x-1)}{dx}+\sin(x-1)\times\frac{d(x-1)}{dx}}$

[Here, Numerator is differentiated by Leibnitz's Rule and denominator by chain rule ] then ,

$\sf\lim{x\to1}\:\dfrac{2(x-1)(x-1)^2\cos(x-1)^2-0}{(x-1)\cos(x-1)+\sin(x-1)\times1}$

$\sf\lim{x\to1}\:\dfrac{2(x-1)(x-1)^2\cos(x-1)^2}{(x-1)\cos(x-1)+\sin(x-1)}$

Again this limit , in the form of 0/0 , After Rearranging terms :

$\sf\lim{x\to1}\:\dfrac{2(x-1)(x-1)^2\cos(x-1)^2}{(x-1)[\cos(x-1)+\frac{\sin(x-1)}{(x-1)}]}$

$\sf\lim{x\to1}\:\dfrac{2\cancel{(x-1)}(x-1)^2\cos(x-1)^2}{\cancel{(x-1)}[\cos(x-1)+\frac{\sin(x-1)}{(x-1)}]}$

$\sf\lim{x\to1}\:\dfrac{2(x-1)^2\cos(x-1)^2}{\cos(x-1)+\frac{\sin(x-1)}{(x-1)}}$

$\sf\green{We\:know\:that}$

$\sf\lim{x\to0}\dfrac{\sin\:x}{x}=1$

Then ,

$\sf\lim{x\to1}\:\dfrac{2(x-1)^2\cos(x-1)^2}{\cos(x-1)+\frac{\sin(x-1)}{(x-1)}}$

put the limit value

$\sf=\dfrac{2\times(1-1)^2\times\cos(1-1)^2}{\cos(1-1)+1}$

$\sf=\dfrac{2\times0\times\cos(0)^2}{\cos(0)+1}$

$\sf=\dfrac{0}{1+1}$

$\sf=\dfrac{0}{2}$

$\sf=0$

Therefore,

$\tt\lim{x\to1}\:\dfrac{\int\limits_{0}^{(x-1)^2}t\cos(t)dt}{(x-1)\sin(x-1)}=0$