Math Problem Statement

lim/x→0 sin x - tan x / x^3

Solution

To solve the limit

limx0sinxtanxx3\lim_{{x \to 0}} \frac{\sin x - \tan x}{x^3}

we'll analyze the expression in the numerator, sinxtanx\sin x - \tan x, and rewrite it to make it easier to evaluate.

Step 1: Rewrite tanx\tan x in terms of sinx\sin x and cosx\cos x

We know that tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}, so we can rewrite the expression as: sinxsinxcosxx3=sinx(11cosx)x3\frac{\sin x - \frac{\sin x}{\cos x}}{x^3} = \frac{\sin x \left(1 - \frac{1}{\cos x}\right)}{x^3}

Step 2: Simplify 11cosx1 - \frac{1}{\cos x}

Using the identity 11cosx=cosx1cosx1 - \frac{1}{\cos x} = \frac{\cos x - 1}{\cos x}, we can rewrite the expression as: sinxcosx1cosxx3=sinx(cosx1)x3cosx\frac{\sin x \cdot \frac{\cos x - 1}{\cos x}}{x^3} = \frac{\sin x \cdot (\cos x - 1)}{x^3 \cdot \cos x}

Step 3: Expand sinx\sin x and cosx1\cos x - 1 using Taylor series

For small xx, we can use the Taylor series expansions: sinxxx36+O(x5)\sin x \approx x - \frac{x^3}{6} + O(x^5) cosx1x22+O(x4)\cos x - 1 \approx -\frac{x^2}{2} + O(x^4)

Substitute these approximations into the expression: (xx36)(x22)x3(1+O(x2))\frac{\left(x - \frac{x^3}{6}\right) \cdot \left(-\frac{x^2}{2}\right)}{x^3 \cdot \left(1 + O(x^2)\right)}

Step 4: Simplify the expression

Multiplying out terms, we get: (x32+x512)x3=x32+x512x3=12+x212\frac{\left(-\frac{x^3}{2} + \frac{x^5}{12}\right)}{x^3} = \frac{-\frac{x^3}{2} + \frac{x^5}{12}}{x^3} = -\frac{1}{2} + \frac{x^2}{12}

Step 5: Take the limit as x0x \to 0

As x0x \to 0, the x212\frac{x^2}{12} term goes to zero, leaving us with: limx0sinxtanxx3=12\lim_{{x \to 0}} \frac{\sin x - \tan x}{x^3} = -\frac{1}{2}

Final Answer

12\boxed{-\frac{1}{2}}

Would you like additional details on Taylor series expansion, or help with any questions about this process?


  1. What is the Taylor series and how does it help in evaluating limits?
  2. How can you approximate sinx\sin x and cosx\cos x for small xx?
  3. Why do higher-order terms vanish when taking limits?
  4. What other methods can be used for limits involving trigonometric functions?
  5. How does Taylor expansion compare with L'Hôpital's Rule for solving limits?

Tip: For limits involving indeterminate forms like 00\frac{0}{0}, Taylor expansions can simplify the problem by approximating functions around the point of interest.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
Taylor Series Expansion

Formulas

lim_{x→0} (sin x - tan x) / x^3
sin x ≈ x - x^3/6 + O(x^5)
cos x - 1 ≈ -x^2/2 + O(x^4)

Theorems

Taylor Series Expansion
Limit Laws

Suitable Grade Level

Grades 11-12