Question

Can anyone solve my integral + limit question?
$\displaystyle\lim_{x\to 1} \dfrac{\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{(x-1)\sin(x-1)}$

Mordetor solve please.

Answer 1

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

Given expression is

$\rm \: \displaystyle\lim_{x\to 1} \dfrac{ \displaystyle\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{(x-1)\sin(x-1)}$

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

So, above expression can be rewritten as

$\rm \: = \: \displaystyle\lim_{x\to 1} \dfrac{ \displaystyle\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{(x-1)^{2} \: \dfrac{sin(x-1)}{x - 1} }$

can be further rewritten as

$\rm \: = \: \displaystyle\lim_{x\to 1} \dfrac{ \displaystyle\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{(x-1)^{2} \: } \times \displaystyle\lim_{x\to 1} \frac{x - 1}{sin(x-1)}$

$\rm \: = \: \displaystyle\lim_{x\to 1} \dfrac{ \displaystyle\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{(x-1)^{2} \: } \times \displaystyle\lim_{x - 1\to 0} \frac{x - 1}{sin(x-1)}$

$\rm \: = \: \displaystyle\lim_{x\to 1} \dfrac{ \displaystyle\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{(x-1)^{2} \: } \times 1$

Now, on applying L Hospital Rule, we get

$\rm \: = \: \displaystyle\lim_{x\to 1} \dfrac{\dfrac{d}{dx} \displaystyle\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{\dfrac{d}{dx} \: (x-1)^{2} \: }$

Using Leibnitz Rule to differentiate under integral sign, we get

$\rm \: = \: \displaystyle\lim_{x\to 1} \frac{\dfrac{d}{dx}{(x - 1)}^{2} \: \times \: [ {(x - 1)}^{2}cos {(x - 1)}^{4}]}{2(x - 1)}$

$\rm \: = \: \displaystyle\lim_{x\to 1} \frac{2(x - 1) \: \times \: [ {(x - 1)}^{2}cos {(x - 1)}^{4}]}{2(x - 1)}$

$\rm \: = \: \displaystyle\lim_{x\to 1} {(x - 1)}^{2} \: cos {(x - 1)}^{4}$

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

$\rm \: = \: 0$

Hence,

$\rm\implies \:\displaystyle\lim_{x\to 1} \dfrac{ \displaystyle\int\limits^{(x-1)^2}_0t\cos(t^2) dt}{(x-1)\sin(x-1)} = 0$

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

Formulae Used :-

$\boxed{ \rm{ \:\displaystyle\lim_{x\to 0} \: \frac{sinx}{x} = 1 \: \: }}$

Leibnitz Rule :-

$\rm \dfrac{d}{dx}\displaystyle\int\limits^{b}_af(t,x)dt =\displaystyle\int\limits^{b}_a\dfrac{ \partial}{ \partial x}f(t,x)dt + \dfrac{d}{dx}b[f(b,x)] - \dfrac{d}{dx}a[f(a,x)]$

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

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{sinx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{tanx}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{log(1 + x)}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {e}^{x} - 1}{x} = 1}\\ \\ \bigstar \: \bf{\displaystyle \rm\lim_{x \to0} \frac{ {a}^{x} - 1}{x} = loga}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$