lim_(x→2) ((1−sin (π/x))/(x^2 −4x+4)) =? lim_(x→2) ((x^3 −8+sin πx)/(2−x))=? lim


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Question Number 161583 by cortano last updated on 19/Dec/21

$\lim_{x \to 2} \frac{1 - \sin \frac{π}{x}}{x^{2} - 4 x + 4} = ?$ $\lim_{x \to 2} \frac{x^{3} - 8 + \sin π x}{2 - x} = ?$ $\lim_{x \to 1} \frac{x^{2} - 1}{\cos (\frac{π}{x + 1})} = ?$
Commented by blackmamba last updated on 20/Dec/21

$(1) \lim_{x \to 2} \frac{\sin (\frac{π}{2}) - \sin (\frac{π}{x})}{{(x - 2)}^{2}}$ $= \lim_{x \to 2} \frac{2 \cos (\frac{π x + 2 π}{4 x}) \sin (\frac{π x - 2 π}{4 x})}{{(x - 2)}^{2}}$ $= \lim_{x \to 2} \frac{2 \sin (\frac{π}{2} - (\frac{π x + 2 π}{4 x})) \sin \frac{π}{4 x} (x - 2)}{{(x - 2)}^{2}}$ $= \lim_{x \to 2} \frac{2 \sin \frac{π}{4 x} (x - 2) \sin \frac{π}{4 x} (x - 2)}{{(x - 2)}^{2}}$ $= \frac{2 π^{2}}{64} = \frac{π^{2}}{32}$
	Answered by Ar Brandon last updated on 20/Dec/21

	$\lim_{x \to 1} \frac{x^{2} - 1}{\cos (\frac{π}{x + 1})} = \lim_{x \to 1} \frac{(x + 1) (x - 1)}{\cos (\frac{π}{x + 1})} = \lim_{x \to 1} \frac{(x + 1) (x - 1)}{\sin (\frac{π}{2} - \frac{π}{x + 1})}$ $\lim_{x \to 1} \frac{(x + 1) (x - 1)}{\sin (\frac{π (x - 1)}{2 (x + 1)})} = \lim_{x \to 1} \frac{2 {(x + 1)}^{2}}{π} \cdot \frac{\frac{π}{2} (\frac{x - 1}{x + 1})}{\sin (\frac{π}{2} (\frac{x - 1}{x + 1}))}$ $= \lim_{x \to 1} \frac{2 {(x + 1)}^{2}}{π} \cdot \lim_{x \to 1} (\frac{\frac{π}{2} (\frac{x - 1}{x + 1})}{\sin (\frac{π}{2} (\frac{x - 1}{x + 1}))}) = \frac{8}{π} \times 1 = \begin{matrix} \frac{8}{π} \end{matrix}$
	Answered by Ar Brandon last updated on 20/Dec/21

	$\lim_{x \to 2} \frac{x^{3} - 8 + \sin π x}{2 - x} = \lim_{x \to 2} \frac{(x - 2) (x^{2} + 2 x + 4)}{2 - x} + \lim_{u \to 0} \frac{\sin (π (2 - u))}{u}$ $= - 12 + \lim_{u \to 0} \frac{\sin (- π u)}{u} = - 12 - \lim_{u \to 0} π \cdot \frac{\sin (π u)}{π u} = - (π + 12)$
	Answered by mathmax by abdo last updated on 20/Dec/21
	$f(x)=((1−sin((π/x)))/(x^2 −4x+4)) changement (π/x)=t give x=(π/t) f(x)=((1−sint)/((π^2 /t^2 )−((4π)/t)+4))=((t^2 (1−sint))/(π^2 −4πt +4t^2 )) (t→(π/2)) =_(t−(π/2)=z) (((z+(π/2))^2 (1−cosz))/(4(z+(π/2))^2 −4π(z+(π/2))+π^2 ))=Ψ(z) (z→0) Ψ(z)∼(z^2 /2)×(((z+(π/2))^2 )/(4(z^2 +πz+(π^2 /4))−4πz−π^2 )) ⇒ Ψ(z)∼((z^2 (z+(π/2))^2 )/(2{4z^2 +4πz+π^2 −4πz−π^2 }))∼(1/8)(z+(π/2))^2 ⇒lim_(z→0) Ψ(z)=(1/8)(π^2 /4)=(π^2 /(32))=limf(x)$
	$f (x) = \frac{1 - \sin (\frac{π}{x})}{x^{2} - 4 x + 4} changement \frac{π}{x} = t give x = \frac{π}{t}$ $f (x) = \frac{1 - sint}{\frac{π^{2}}{t^{2}} - \frac{4 π}{t} + 4} = \frac{t^{2} (1 - sint)}{π^{2} - 4 π t + {4 t}^{2}} (t \to \frac{π}{2})$ $=_{t - \frac{π}{2} = z} \frac{{(z + \frac{π}{2})}^{2} (1 - cosz)}{4 {(z + \frac{π}{2})}^{2} - 4 π (z + \frac{π}{2}) + π^{2}} = Ψ (z) (z \to 0)$ $Ψ (z) \sim \frac{z^{2}}{2} \times \frac{{(z + \frac{π}{2})}^{2}}{4 (z^{2} + π z + \frac{π^{2}}{4}) - 4 π z - π^{2}} \Rightarrow$ $Ψ (z) \sim \frac{z^{2} {(z + \frac{π}{2})}^{2}}{2 {{4 z}^{2} + 4 π z + π^{2} - 4 π z - π^{2}}} \sim \frac{1}{8} {(z + \frac{π}{2})}^{2}$ $\Rightarrow \lim_{z \to 0} Ψ (z) = \frac{1}{8} \frac{π^{2}}{4} = \frac{π^{2}}{32} = limf (x)$
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