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Question: $\lim_{x \to \pi/2} \frac{sin (\pi \csc^2 x)}{(\pi-2x)^2}$...

lim⁡x→π/2sin(πcsc⁡2x)(π−2x)2\lim_{x \to \pi/2} \frac{sin (\pi \csc^2 x)}{(\pi-2x)^2}

Answer

−π4-\frac{\pi}{4}

Explanation

Solution

To evaluate the limit lim⁡x→π/2sin⁡(πcsc⁡2x)(π−2x)2\lim_{x \to \pi/2} \frac{\sin (\pi \csc^2 x)}{(\pi-2x)^2}:

1. Identify Indeterminate Form:

As x→π/2x \to \pi/2:

Numerator: sin⁡(πcsc⁡2x)→sin⁡(π⋅12)=sin⁡(π)=0\sin (\pi \csc^2 x) \to \sin (\pi \cdot 1^2) = \sin(\pi) = 0.

Denominator: (π−2x)2→(π−2(π/2))2=(π−π)2=02=0(\pi-2x)^2 \to (\pi - 2(\pi/2))^2 = (\pi-\pi)^2 = 0^2 = 0.

The limit is of the indeterminate form 0/00/0.

2. Transform the Numerator using Trigonometric Identities:

We use the identity sin⁡A=sin⁡(π−A)\sin A = \sin(\pi - A).

So, sin⁡(πcsc⁡2x)=sin⁡(π−πcsc⁡2x)\sin(\pi \csc^2 x) = \sin(\pi - \pi \csc^2 x).

Factor out π\pi from the argument: π(1−csc⁡2x)\pi(1 - \csc^2 x).

Recall the identity 1+cot⁡2x=csc⁡2x1 + \cot^2 x = \csc^2 x, which implies 1−csc⁡2x=−cot⁡2x1 - \csc^2 x = -\cot^2 x.

Substitute this into the argument: π(−cot⁡2x)=−πcot⁡2x\pi(-\cot^2 x) = -\pi \cot^2 x.

Therefore, the numerator becomes sin⁡(−πcot⁡2x)\sin(-\pi \cot^2 x).

Since sin⁡(−A)=−sin⁡A\sin(-A) = -\sin A, we have sin⁡(−πcot⁡2x)=−sin⁡(πcot⁡2x)\sin(-\pi \cot^2 x) = -\sin(\pi \cot^2 x).

The limit expression now is:

lim⁡x→π/2−sin⁡(πcot⁡2x)(π−2x)2\lim_{x \to \pi/2} \frac{-\sin(\pi \cot^2 x)}{(\pi-2x)^2}

3. Apply Standard Limit lim⁡y→0sin⁡yy=1\lim_{y \to 0} \frac{\sin y}{y} = 1:

As x→π/2x \to \pi/2, cot⁡x→cot⁡(π/2)=0\cot x \to \cot(\pi/2) = 0. So, πcot⁡2x→0\pi \cot^2 x \to 0.

Let y=πcot⁡2xy = \pi \cot^2 x. We can multiply and divide by yy to use the standard limit:

lim⁡x→π/2(−sin⁡(πcot⁡2x)πcot⁡2x⋅πcot⁡2x(π−2x)2)\lim_{x \to \pi/2} \left( -\frac{\sin(\pi \cot^2 x)}{\pi \cot^2 x} \cdot \frac{\pi \cot^2 x}{(\pi-2x)^2} \right)

The first part, lim⁡x→π/2−sin⁡(πcot⁡2x)πcot⁡2x\lim_{x \to \pi/2} \frac{-\sin(\pi \cot^2 x)}{\pi \cot^2 x}, evaluates to −1-1 as y→0y \to 0.

So, the limit simplifies to:

−1⋅lim⁡x→π/2πcot⁡2x(π−2x)2-1 \cdot \lim_{x \to \pi/2} \frac{\pi \cot^2 x}{(\pi-2x)^2}

4. Simplify and Use Substitution:

Rewrite cot⁡2x\cot^2 x as cos⁡2xsin⁡2x\frac{\cos^2 x}{\sin^2 x}:

−πlim⁡x→π/2cos⁡2x(π−2x)2sin⁡2x-\pi \lim_{x \to \pi/2} \frac{\cos^2 x}{(\pi-2x)^2 \sin^2 x}

As x→π/2x \to \pi/2, sin⁡2x→sin⁡2(π/2)=12=1\sin^2 x \to \sin^2(\pi/2) = 1^2 = 1.

So, the expression becomes:

−πlim⁡x→π/2cos⁡2x(π−2x)2-\pi \lim_{x \to \pi/2} \frac{\cos^2 x}{(\pi-2x)^2}

Let x=π/2−hx = \pi/2 - h. As x→π/2x \to \pi/2, h→0h \to 0.

Substitute x=π/2−hx = \pi/2 - h:

  • cos⁡x=cos⁡(π/2−h)=sin⁡h\cos x = \cos(\pi/2 - h) = \sin h
  • π−2x=π−2(π/2−h)=π−π+2h=2h\pi - 2x = \pi - 2(\pi/2 - h) = \pi - \pi + 2h = 2h

Substitute these into the limit expression:

−πlim⁡h→0(sin⁡h)2(2h)2-\pi \lim_{h \to 0} \frac{(\sin h)^2}{(2h)^2}

−πlim⁡h→0sin⁡2h4h2-\pi \lim_{h \to 0} \frac{\sin^2 h}{4h^2}

−πlim⁡h→014(sin⁡hh)2-\pi \lim_{h \to 0} \frac{1}{4} \left( \frac{\sin h}{h} \right)^2

Using the standard limit lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1:

−π⋅14(1)2=−π4-\pi \cdot \frac{1}{4} (1)^2 = -\frac{\pi}{4}

The final answer is −π4-\frac{\pi}{4}.

Explanation of the solution:

The limit is of the form 0/00/0. Transform the numerator sin⁡(πcsc⁡2x)\sin(\pi \csc^2 x) to −sin⁡(πcot⁡2x)-\sin(\pi \cot^2 x) using sin⁡A=sin⁡(π−A)\sin A = \sin(\pi-A) and 1−csc⁡2x=−cot⁡2x1-\csc^2 x = -\cot^2 x. Apply the standard limit lim⁡y→0sin⁡yy=1\lim_{y \to 0} \frac{\sin y}{y}=1. The problem reduces to evaluating lim⁡x→π/2−πcot⁡2x(π−2x)2\lim_{x \to \pi/2} \frac{-\pi \cot^2 x}{(\pi-2x)^2}. Convert cot⁡2x\cot^2 x to cos⁡2xsin⁡2x\frac{\cos^2 x}{\sin^2 x}. Substitute x=π/2−hx = \pi/2 - h and simplify using cos⁡(π/2−h)=sin⁡h\cos(\pi/2 - h) = \sin h and π−2x=2h\pi-2x = 2h. The limit becomes −πlim⁡h→0sin⁡2h4h2-\pi \lim_{h \to 0} \frac{\sin^2 h}{4h^2}, which evaluates to −π/4-\pi/4.