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Question: If $L = \lim_{x\to\frac{\pi}{2}} \frac{\sqrt{1 + \cot x} - \sqrt{1 + \cos x}}{(\pi - 2x)^3}$, then ...

If

L=lim⁡x→π21+cot⁡x−1+cos⁡x(π−2x)3L = \lim_{x\to\frac{\pi}{2}} \frac{\sqrt{1 + \cot x} - \sqrt{1 + \cos x}}{(\pi - 2x)^3}, then 96 L96 \ L is _______ .

Answer

3

Explanation

Solution

To evaluate the limit L=lim⁡x→π21+cot⁡x−1+cos⁡x(π−2x)3L = \lim_{x\to\frac{\pi}{2}} \frac{\sqrt{1 + \cot x} - \sqrt{1 + \cos x}}{(\pi - 2x)^3}, we first observe that as x→π2x \to \frac{\pi}{2}, the expression is of the indeterminate form 00\frac{0}{0}.

Let's make a substitution to simplify the limit. Let x=π2+hx = \frac{\pi}{2} + h. As x→π2x \to \frac{\pi}{2}, h→0h \to 0.

Now, substitute x=π2+hx = \frac{\pi}{2} + h into the expression:

  1. Denominator: (π−2x)3=(π−2(π2+h))3=(π−π−2h)3=(−2h)3=−8h3(\pi - 2x)^3 = (\pi - 2(\frac{\pi}{2} + h))^3 = (\pi - \pi - 2h)^3 = (-2h)^3 = -8h^3.

  2. Numerator: cot⁡x=cot⁡(π2+h)=−tan⁡h\cot x = \cot(\frac{\pi}{2} + h) = -\tan h. cos⁡x=cos⁡(π2+h)=−sin⁡h\cos x = \cos(\frac{\pi}{2} + h) = -\sin h. So, the numerator becomes 1−tan⁡h−1−sin⁡h\sqrt{1 - \tan h} - \sqrt{1 - \sin h}.

The limit expression transforms to: L=lim⁡h→01−tan⁡h−1−sin⁡h−8h3L = \lim_{h\to 0} \frac{\sqrt{1 - \tan h} - \sqrt{1 - \sin h}}{-8h^3}

To simplify the numerator, multiply and divide by its conjugate: L=lim⁡h→0(1−tan⁡h−1−sin⁡h)(1−tan⁡h+1−sin⁡h)−8h3(1−tan⁡h+1−sin⁡h)L = \lim_{h\to 0} \frac{(\sqrt{1 - \tan h} - \sqrt{1 - \sin h})(\sqrt{1 - \tan h} + \sqrt{1 - \sin h})}{-8h^3(\sqrt{1 - \tan h} + \sqrt{1 - \sin h})} L=lim⁡h→0(1−tan⁡h)−(1−sin⁡h)−8h3(1−tan⁡h+1−sin⁡h)L = \lim_{h\to 0} \frac{(1 - \tan h) - (1 - \sin h)}{-8h^3(\sqrt{1 - \tan h} + \sqrt{1 - \sin h})} L=lim⁡h→0sin⁡h−tan⁡h−8h3(1−tan⁡h+1−sin⁡h)L = \lim_{h\to 0} \frac{\sin h - \tan h}{-8h^3(\sqrt{1 - \tan h} + \sqrt{1 - \sin h})}

As h→0h \to 0, the term 1−tan⁡h+1−sin⁡h\sqrt{1 - \tan h} + \sqrt{1 - \sin h} approaches 1−0+1−0=1+1=2\sqrt{1 - 0} + \sqrt{1 - 0} = 1 + 1 = 2. So, we can write: L=1−8×2lim⁡h→0sin⁡h−tan⁡hh3L = \frac{1}{-8 \times 2} \lim_{h\to 0} \frac{\sin h - \tan h}{h^3} L=−116lim⁡h→0sin⁡h−sin⁡hcos⁡hh3L = -\frac{1}{16} \lim_{h\to 0} \frac{\sin h - \frac{\sin h}{\cos h}}{h^3} L=−116lim⁡h→0sin⁡h(1−1cos⁡h)h3L = -\frac{1}{16} \lim_{h\to 0} \frac{\sin h (1 - \frac{1}{\cos h})}{h^3} L=−116lim⁡h→0sin⁡h(cos⁡h−1cos⁡h)h3L = -\frac{1}{16} \lim_{h\to 0} \frac{\sin h (\frac{\cos h - 1}{\cos h})}{h^3}

Now, rearrange the terms to use standard limits: L=−116lim⁡h→0(sin⁡hh×cos⁡h−1h2×1cos⁡h)L = -\frac{1}{16} \lim_{h\to 0} \left( \frac{\sin h}{h} \times \frac{\cos h - 1}{h^2} \times \frac{1}{\cos h} \right)

We know the following standard limits:

  1. lim⁡h→0sin⁡hh=1\lim_{h\to 0} \frac{\sin h}{h} = 1
  2. lim⁡h→0cos⁡h−1h2=−12\lim_{h\to 0} \frac{\cos h - 1}{h^2} = -\frac{1}{2}
  3. lim⁡h→01cos⁡h=1cos⁡0=11=1\lim_{h\to 0} \frac{1}{\cos h} = \frac{1}{\cos 0} = \frac{1}{1} = 1

Substitute these values into the expression for LL: L=−116×(1)×(−12)×(1)L = -\frac{1}{16} \times (1) \times (-\frac{1}{2}) \times (1) L=−116×(−12)L = -\frac{1}{16} \times (-\frac{1}{2}) L=132L = \frac{1}{32}

The question asks for the value of 96L96L. 96L=96×132=396L = 96 \times \frac{1}{32} = 3.

The final answer is 3\boxed{3}.

Explanation of the solution:

  1. Substitute x=π2+hx = \frac{\pi}{2} + h to transform the limit as h→0h \to 0.
  2. Simplify trigonometric terms: cot⁡(π2+h)=−tan⁡h\cot(\frac{\pi}{2} + h) = -\tan h and cos⁡(π2+h)=−sin⁡h\cos(\frac{\pi}{2} + h) = -\sin h.
  3. Rationalize the numerator by multiplying by its conjugate.
  4. Factor out sin⁡h\sin h from the numerator and rearrange terms to identify standard limits: lim⁡h→0sin⁡hh\lim_{h\to 0} \frac{\sin h}{h}, lim⁡h→0cos⁡h−1h2\lim_{h\to 0} \frac{\cos h - 1}{h^2}, and lim⁡h→01cos⁡h\lim_{h\to 0} \frac{1}{\cos h}.
  5. Substitute the values of these standard limits to find LL.
  6. Calculate 96L96L.