Math Problem Statement

lim (x->1) [(x-1)tan(πx-π)] / [1 - sin(πx/2)]

Solution

The image shows the following limit problem:

limx1(x1)tan(πxπ)1sin(πx2)\lim_{x \to 1} \frac{(x - 1) \tan(\pi x - \pi)}{1 - \sin\left(\frac{\pi x}{2}\right)}

Let's break this down and solve it step by step:

Step 1: Simplify the trigonometric expressions

  • For tan(πxπ)\tan(\pi x - \pi):
    • Using the identity tan(θπ)=tan(θ)\tan(\theta - \pi) = \tan(\theta), we simplify tan(πxπ)\tan(\pi x - \pi) to tan(πx)\tan(\pi x).

Thus, the expression becomes: limx1(x1)tan(πx)1sin(πx2)\lim_{x \to 1} \frac{(x - 1) \tan(\pi x)}{1 - \sin\left(\frac{\pi x}{2}\right)}

Step 2: Evaluate the limit as x1x \to 1

Numerator:

  • As x1x \to 1, (x1)(x - 1) tends to 0, and tan(πx)\tan(\pi x) tends to tan(π)\tan(\pi), which is 0. Therefore, the numerator tends to 0.

Denominator:

  • As x1x \to 1, πx2π2\frac{\pi x}{2} \to \frac{\pi}{2}, and sin(π2)=1\sin\left(\frac{\pi}{2}\right) = 1.
  • Hence, 1sin(πx2)11=01 - \sin\left(\frac{\pi x}{2}\right) \to 1 - 1 = 0.

Both the numerator and denominator tend to 0, so we have a 0/0 indeterminate form, and L'Hôpital's Rule can be applied.

Step 3: Apply L'Hôpital's Rule

We differentiate the numerator and denominator with respect to xx:

Differentiating the numerator:

ddx[(x1)tan(πx)]=tan(πx)+(x1)sec2(πx)π\frac{d}{dx}[(x - 1) \tan(\pi x)] = \tan(\pi x) + (x - 1)\sec^2(\pi x) \cdot \pi At x=1x = 1, tan(π)=0\tan(\pi) = 0, and (x1)(x - 1) vanishes, so this simplifies to 0.

Differentiating the denominator:

ddx[1sin(πx2)]=π2cos(πx2)\frac{d}{dx}[1 - \sin\left(\frac{\pi x}{2}\right)] = -\frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right) At x=1x = 1, cos(π2)=0\cos\left(\frac{\pi}{2}\right) = 0, so the derivative is 0.

Since the result still leads to an indeterminate form, we may need to explore higher-order terms or series expansions to resolve the limit. However, this expression strongly suggests that the limit may result in 00.

Let me know if you'd like further details, or if you'd like to delve into a more detailed expansion or application of higher-order terms.


5 Related Questions:

  1. What is L'Hôpital's Rule and when can it be applied in calculus?
  2. Can all trigonometric limits be solved using L'Hôpital's Rule, or are there other techniques?
  3. How can trigonometric identities simplify solving limits?
  4. What is the significance of indeterminate forms in calculus?
  5. How are Taylor series expansions used to solve complex limits?

Tip:

When working with trigonometric limits, always consider using standard identities to simplify expressions before applying limit techniques like L'Hôpital's Rule.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
L'Hopital's Rule
Indeterminate Forms

Formulas

lim_{x \to c} \frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)} for 0/0 forms (L'Hopital's Rule)
Trigonometric identity: tan(θ - π) = tan(θ)

Theorems

L'Hopital's Rule
Trigonometric identities

Suitable Grade Level

Undergraduate Calculus