There is a problem that is basic to solve with L\"hospital yet we are required to settle it there is no it, yet I might not discover the answer. X* sinx part is particularly confusing me, because other instances I have solved did not incorporate such part.

$$\\lim_x \\to π/2\\frac2x\\sin(x) - π\\cos x$$




You are watching: Lim x → (π/2)+ cos(x) 1 − sin(x)

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Define $f(x) = 2x\\sin x, g(x) = \\cos x.$ The expression equals

$$\\fracf(x) - f(\\pi/2)g(x) - g(\\pi/2) = \\dfrac \\dfracf(x) - f(\\pi/2)x-\\pi /2 \\dfracg(x) - g(\\pi/2)x-\\pi/2.$$

By definition of the derivative, the last expression $\\to f\"(\\pi/2)/g\"(\\pi/2),$ and simple computation shows this equates to $-2.$ (And no, we did not usage L\"Hopital.)


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Let $t=x-\\pi/2$. Then, us have

$$\\beginalign\\lim_x\\to \\pi/2\\frac2x\\sin(x)-\\pi\\cos(x)&=\\lim_t\\to 0\\frac2(t+\\pi/2)\\sin(t+\\pi/2)-\\pi\\cos(t+\\pi/2)\\\\\\\\&=\\lim_t\\to 0\\frac\\pi(1-\\cos(t))-2t\\cos(t)\\sin(t)\\\\\\\\&=\\lim_t\\to 0\\left(2\\pi \\frac\\sin^2(t/2)\\sin(t)-2\\,\\frac\\cos(t)\\sin(t)/t\\right)\\\\\\\\&=\\lim_t\\to 0\\left(\\pi \\frac\\sin(t/2)\\cos(t/2)-2\\,\\frac\\cos(t)\\sin(t)/t\\right)\\\\\\\\&=-2\\endalign$$


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$$\\beginalign\\lim_x \\to π/2\\frac2x\\sin(x) - π\\cos x = &\\lim_x \\to π/2\\frac2x\\sin(x) - π\\sin \\left(\\dfrac\\pi2 -x\\right) &\\\\= &\\lim_x \\to π/2\\frac2x\\sin(x) - π\\left(\\dfrac\\pi2 -x\\right)\\dfrac\\sin \\left(\\dfrac\\pi2 -x\\right)\\dfrac\\pi2 -x &\\\\=&2\\lim_x \\to π/2\\fracx\\sin(x) - \\dfracπ2\\left(\\dfrac\\pi2 -x\\right)&\\\\=&2\\lim_x \\to π/2\\fracx\\sin(x) - x - \\left(\\dfracπ2 -x \\right)\\left(\\dfrac\\pi2 -x\\right)&\\\\=&-2 +2\\lim_x \\to π/2\\fracx(\\sin(x) - 1) \\left(\\dfrac\\pi2 -x\\right) &\\\\=&-2+\\pi\\lim_x \\to π/2\\frac\\cos\\left(\\dfrac\\pi2 - x\\right) - 1 \\left(\\dfrac\\pi2 -x\\right)&\\\\=&-2 +\\pi\\lim_x \\to π/2-\\frac\\sin^2\\left(\\dfrac\\pi4 - \\dfrac x2\\right) \\left(\\dfrac\\pi2 -x\\right) = -2\\endalign$$




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Substitute $y=\\frac\\pi 2 - x$. Then

$$\\lim_x \\to π/2\\frac2x\\sin x - π\\cos x = \\lim_y\\to0\\frac2(\\frac\\pi 2-y)\\cos y - π\\sin y$$and the function under the limit is$$\\frac2(\\frac\\pi 2-y)\\cos y - π\\sin y= \\frac(\\pi-2y)\\cos y - π\\sin y \\\\= \\frac\\pi(\\cos y - 1)-2y\\cos y\\sin y \\\\= \\pi\\frac\\cos y-1\\sin y-\\frac2y\\sin y\\cos y \\\\= \\pi\\frac\\colorred\\cos y-\\colorblue1\\sin y-2\\frac y\\sin y\\cos y \\\\= \\pi\\frac\\colorred\\cos(2\\cdot\\frac y2)-\\colorblue1\\sin(2\\cdot\\frac y2)-2\\frac y\\sin y\\cos y \\\\= \\pi\\frac\\colorred(\\cos^2\\frac y2-\\sin^2\\frac y2)-\\colorblue(\\sin^2\\frac y2+\\cos^2\\frac y2)2\\sin\\frac y2\\cdot\\cos\\frac y2-2\\frac y\\sin y\\cos y \\\\= \\pi\\frac-2\\sin^2\\frac y22\\sin\\frac y2\\cdot\\cos\\frac y2-2\\frac y\\sin y\\cos y \\\\= -\\pi\\frac\\sin\\frac y2\\cos\\frac y2-2\\frac y\\sin y\\cos y \\\\= -\\pi\\tan\\frac y2-2\\frac y\\sin y\\cos y$$We know (...?) that$$\\tan 0 = 0$$$$\\cos 0 = 1$$and$$\\lim_y\\to 0\\frac y\\sin y = 1$$so the border sought is$$\\lim_y\\to 0 \\left(-\\pi\\tan\\frac y2-2\\frac y\\sin y\\cos y\\right) = -\\pi \\cdot 0 - 2\\cdot 1 \\cdot 1 = \\boxed-2$$