Question

$\large \displaystyle\lim_{ x \rightarrow 1 } \left( \frac{ \displaystyle\int_{ 0 }^{ (x-1) ^ { 2 } } \cos ( t ) d t }{ (x-1) \sin ( x-1 ) } \right)$

A challenge for all brainly stars and moderators...​

Answer 1

$\bf{Hello\:\: Mate\:\: !!!!!$

$\bf{Answer\:\::}$

$\bf{\lim _{x\to \:1}\left(\dfrac{\int _0^{\left(x-1\right)^2}\cos \left(t\right)dt}{\left(x-1\right)\sin \left(x-1\right)}\right)=0}$

$\bf{Explanation\:\::}$

$\bf{Given\:\::}$

$\bf{\lim _{x\to \:1}\left(\dfrac{\int _0^{\left(x-1\right)^2}\cos \left(t\right)dt}{\left(x-1\right)\sin \left(x-1\right)}\right)}$

$\rule{250}{6}$

$\bf{\int _0^{\left(x-1\right)^2}\cos \left(t\right)dt=\sin \left(\left(x-1\right)^2\right)}$

$=\bf{\lim _{x\to \:1}\left(\dfrac{\sin \left(\left(x-1\right)^2\right)}{\left(x-1\right)\sin \left(x-1\right)}\right)}$

$\bf{\lim _{x\to \:1}\left(\dfrac{\sin \left(\left(x-1\right)^2\right)}{\left(x-1\right)\sin \left(x-1\right)}\right)=0}$

$\bf{SO\:\:OUR\:\:ANSWER\:\:TO\:\:QUESTION :}$

$\bf{\lim _{x\to \:1}\left(\dfrac{\int _0^{\left(x-1\right)^2}\cos \left(t\right)dt}{\left(x-1\right)\sin \left(x-1\right)}\right)=0}$

Answer 2

SOLUTION:-

m-10, a= 20, b = 0.8 The period under simple

harmonic motion is 4 seconds.42

x-lcos (t) dt =0

limx+1 (x - 1) sin(x-; 1)

Given:-

rlx-1)2 cos (t) dt

(x-1) sin (x 1)

limx1d = ae 2 cos4m

2 d 20e 201 cos _(0.8)

4(10F0.64

d 200OSti 2010 cos

=400

ANSWER-✔️............