Q lim x tends to pi/2 1+cos2x / (pi-2x)2

Given, limx→π21+cos2x(π−2x)2=limx→π22cos2x4(π2−x)2=12limx→π2cos2x(π2−x)2=12limx→π2[sin(π2−x)]2(π2−x)2=12limy→0(sinyy)2 where (π2−x)=y=>y→0 =12(limy→0sinyy)2=12×12 (Since limy→0sinyy=1 )=12

Q. lim x tends to pi/2

1+cos2x / (pi-2x)²

\begin{array}{l} Given, \lim_{x \to \frac{π}{2}} \frac{1 + \cos 2 x}{{(π - 2 x)}^{2}} \\ = \lim_{x \to \frac{π}{2}} \frac{2 \cos^{2} x}{4 {(\frac{π}{2} - x)}^{2}} \\ = \frac{1}{2} \lim_{x \to \frac{π}{2}} \frac{\cos^{2} x}{{(\frac{π}{2} - x)}^{2}} \\ = \frac{1}{2} \lim_{x \to \frac{π}{2}} \frac{{[\sin (\frac{π}{2} - x)]}^{2}}{{(\frac{π}{2} - x)}^{2}} \\ = \frac{1}{2} \lim_{y \to 0} {(\frac{\sin y}{y})}^{2} where (\frac{π}{2} - x) = y = > y \to 0. \\ = \frac{1}{2} {(\lim_{y \to 0} \frac{\sin y}{y})}^{2} \\ = \frac{1}{2} \times 1^{2} (Since \lim_{y \to 0} \frac{\sin y}{y} = 1) \\ = \frac{1}{2} . \end{array}

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Q. lim x tends to pi/2 1+cos2x / (pi-2x)2

Q. lim x tends to pi/2

1+cos2x / (pi-2x)²