Math Problem Statement

Вычислить предел \( \lim_{x \to \frac{\pi}{2}} \frac{\frac{2}{\cos x} - \sin x}{x - \frac{\pi}{2}} \) используя правило Лопиталя.

Solution

To evaluate the given limit using L'Hôpital's Rule, let us proceed step by step.

The problem is:

limxπ2(2cosxsinxxπ2)\lim_{x \to \frac{\pi}{2}} \left( \frac{\frac{2}{\cos x} - \sin x}{x - \frac{\pi}{2}} \right)


Step 1: Check the indeterminate form

As xπ2x \to \frac{\pi}{2}, evaluate each term:

  1. cos(π2)=0\cos\left(\frac{\pi}{2}\right) = 0, so 2cosx\frac{2}{\cos x} \to \infty.
  2. sin(π2)=1\sin\left(\frac{\pi}{2}\right) = 1, so 2cosxsinx1=\frac{2}{\cos x} - \sin x \to \infty - 1 = \infty.
  3. The denominator xπ20x - \frac{\pi}{2} \to 0.

Thus, the limit is of the form 0\frac{\infty}{0}. This indicates a singularity, and L'Hôpital's Rule applies.


Step 2: Differentiate numerator and denominator

To apply L'Hôpital's Rule, differentiate the numerator N(x)N(x) and denominator D(x)D(x) with respect to xx.

  • Numerator: N(x)=2cosxsinxN(x) = \frac{2}{\cos x} - \sin x

Differentiate N(x)N(x) term by term:

  1. For 2cosx\frac{2}{\cos x}, rewrite as 2secx2 \sec x. Its derivative is: ddx[2secx]=2secxtanx\frac{d}{dx} [2 \sec x] = 2 \sec x \tan x
  2. For sinx\sin x, its derivative is: ddx[sinx]=cosx\frac{d}{dx} [\sin x] = \cos x

Thus, the derivative of the numerator is: N(x)=2secxtanxcosxN'(x) = 2 \sec x \tan x - \cos x

  • Denominator: D(x)=xπ2D(x) = x - \frac{\pi}{2}

The derivative of the denominator is: D(x)=1D'(x) = 1


Step 3: Apply L'Hôpital's Rule

Now, apply L'Hôpital's Rule:

limxπ2N(x)D(x)=limxπ2N(x)D(x)\lim_{x \to \frac{\pi}{2}} \frac{N(x)}{D(x)} = \lim_{x \to \frac{\pi}{2}} \frac{N'(x)}{D'(x)}

Substitute N(x)N'(x) and D(x)D'(x):

limxπ22secxtanxcosx1\lim_{x \to \frac{\pi}{2}} \frac{2 \sec x \tan x - \cos x}{1}


Step 4: Evaluate the limit as xπ2x \to \frac{\pi}{2}

As xπ2x \to \frac{\pi}{2}:

  1. sec(π2)=1cos(π2)\sec\left(\frac{\pi}{2}\right) = \frac{1}{\cos\left(\frac{\pi}{2}\right)} \to \infty
  2. tan(π2)=sin(π2)cos(π2)\tan\left(\frac{\pi}{2}\right) = \frac{\sin\left(\frac{\pi}{2}\right)}{\cos\left(\frac{\pi}{2}\right)} \to \infty
  3. cos(π2)=0\cos\left(\frac{\pi}{2}\right) = 0

Thus, the term 2secxtanx2 \sec x \tan x \to \infty, and the other term cosx0-\cos x \to 0.

Hence, the overall limit is infinite:

limxπ2(2cosxsinxxπ2)=\lim_{x \to \frac{\pi}{2}} \left( \frac{\frac{2}{\cos x} - \sin x}{x - \frac{\pi}{2}} \right) = \infty


Final Answer:

The limit diverges to infinity.


Do you want me to explain any step in more detail or clarify any concept?

Here are 5 related questions to deepen your understanding:

  1. What is L'Hôpital's Rule, and when can it be applied?
  2. How do you differentiate trigonometric functions like secx\sec x and tanx\tan x?
  3. What are common indeterminate forms in limits?
  4. How do you handle singularities in limits?
  5. How can you analyze the behavior of trigonometric functions near critical points like x=π2x = \frac{\pi}{2}?

Tip: Always verify that the original limit is in an indeterminate form before applying L'Hôpital's Rule.

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Math Problem Analysis

Mathematical Concepts

Limits
L'Hôpital's Rule
Trigonometric Functions

Formulas

L'Hôpital's Rule: \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) (if indeterminate form exists)
Derivative of \( \sec x \): \( \frac{d}{dx}[\sec x] = \sec x \tan x \)
Derivative of \( \sin x \): \( \frac{d}{dx}[\sin x] = \cos x \)

Theorems

L'Hôpital's Rule

Suitable Grade Level

Undergraduate Calculus