Question

$\displaystyle\rm \red{the \: value \: of \: \lim_{x \to a} \frac{1}{ {x}^{2} - {a}^{2} } \int_{0}^x \sin( {t}^{2} ) dx,a \ne0 \: is }$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle\lim_{x \to a} \frac{1}{ {x}^{2} - {a}^{2} } \int_{0}^x \sin( {t}^{2} ) dt$

$\displaystyle = \lim_{x \to a} \frac{ \displaystyle\int_{0}^x \sin( {t}^{2} ) dt}{ {x}^{2} - {a}^{2} }$

Using l'hospital's rule,

$\displaystyle = \lim_{x \to a} \frac{ \sin( {x}^{2} ) \cdot \dfrac{d}{dx}(x) - 0}{ 2x }$

$\displaystyle = \lim_{x \to a} \frac{ \sin( {x}^{2} ) \cdot 1}{ 2x }$

$\displaystyle = \lim_{x \to a} \frac{ \sin( {x}^{2} ) }{ 2x }$

$= \dfrac{ \sin( {a}^{2} ) }{ 2a }$

Answer 2

REQUIRED ANSWER

$\displaystyle\lim_{x \to a} \frac{1}{ {x}^{2} - {a}^{2} } \int_{0}^x \sin( {t}^{2} ) dtx</p><p>$

$\displaystyle = \lim_{x \to a} \frac{ \displaystyle\int_{0}^x \sin( {t}^{2} ) dt}{ {x}^{2} - {a}^{2} }=x$

Using l'hospital's rule,

$\displaystyle = \lim_{x \to a} \frac{ \sin( {x}^{2} ) \cdot \dfrac{d}{dx}(x) - 0}{ 2x }=x$

$</p><p>\displaystyle = \lim_{x \to a} \frac{ \sin( {x}^{2} ) \cdot 1}{ 2x }=x</p><p>$

$</p><p>\displaystyle = \lim_{x \to a} \frac{ \sin( {x}^{2} ) }{ 2x }=x</p><p>$

$= \dfrac{ \sin( {a}^{2} ) }{ 2a }$

Hope it helps