Math Problem Statement

Solve this

Solve the limit problem lim (x -> π/2) (1 - sin x) / (π/2 - x)^2

Solution

The given limit problem is:

limxπ21sinx(π2x)2\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}

Step 1: Substituting x=π2x = \frac{\pi}{2}

First, check the form of the limit by directly substituting x=π2x = \frac{\pi}{2}:

  • sin(π2)=1\sin\left( \frac{\pi}{2} \right) = 1,
  • π2π2=0\frac{\pi}{2} - \frac{\pi}{2} = 0.

Substituting into the expression gives:

1sin(π2)(π2π2)2=1102=00\frac{1 - \sin\left( \frac{\pi}{2} \right)}{\left( \frac{\pi}{2} - \frac{\pi}{2} \right)^2} = \frac{1 - 1}{0^2} = \frac{0}{0}

This is an indeterminate form 00\frac{0}{0}, so we can apply L'Hôpital's Rule.

Step 2: Applying L'Hôpital's Rule

L'Hôpital's Rule states that if a limit results in an indeterminate form 00\frac{0}{0}, we can differentiate the numerator and the denominator with respect to xx, and then re-evaluate the limit.

The expression is:

limxπ21sinx(π2x)2\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}

Differentiate the numerator:

The derivative of 1sinx1 - \sin x with respect to xx is:

cosx-\cos x

Differentiate the denominator:

The derivative of (π2x)2\left( \frac{\pi}{2} - x \right)^2 with respect to xx is:

2(π2x)(1)=2(π2x)2\left( \frac{\pi}{2} - x \right) \cdot (-1) = -2 \left( \frac{\pi}{2} - x \right)

So, applying L'Hôpital's Rule gives:

limxπ2cosx2(π2x)\lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-2 \left( \frac{\pi}{2} - x \right)}

Simplifying this:

limxπ2cosx2(π2x)\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2 \left( \frac{\pi}{2} - x \right)}

Step 3: Evaluating the Limit

Now substitute x=π2x = \frac{\pi}{2} into the expression. Since cos(π2)=0\cos \left( \frac{\pi}{2} \right) = 0, we get:

limxπ2cosx2(π2x)=00\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{2 \left( \frac{\pi}{2} - x \right)} = \frac{0}{0}

Again, this results in an indeterminate form, so we apply L'Hôpital's Rule a second time.

Step 4: Second Application of L'Hôpital's Rule

Differentiate the numerator and denominator again.

Numerator:

The derivative of cosx\cos x is:

sinx-\sin x

Denominator:

The derivative of 2(π2x)2 \left( \frac{\pi}{2} - x \right) is:

2-2

So the new limit is:

limxπ2sinx2=limxπ2sinx2\lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{\sin x}{2}

Now substitute x=π2x = \frac{\pi}{2} into this expression:

sin(π2)2=12\frac{\sin \left( \frac{\pi}{2} \right)}{2} = \frac{1}{2}

Final Answer:

limxπ21sinx(π2x)2=12\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2} = \frac{1}{2}

Let me know if you have any questions or need more details.


Here are 5 related questions to expand your understanding:

  1. How does L'Hôpital's Rule help in solving indeterminate forms?
  2. What other techniques can be used to solve limits besides L'Hôpital's Rule?
  3. Can the limit be evaluated using series expansions around x=π2x = \frac{\pi}{2}?
  4. What are the conditions under which L'Hôpital's Rule can be applied?
  5. What would happen if we tried to solve this limit using a Taylor series expansion?

Tip: Always check if a limit is indeterminate before applying L'Hôpital’s Rule—other methods like factorization or series expansion might be simpler!

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometry
Calculus
L'Hopital's Rule

Formulas

L'Hopital's Rule
Derivative of sin x
Derivative of a power function

Theorems

L'Hopital's Rule

Suitable Grade Level

Grades 11-12 or introductory college calculus