Math Problem Statement

find the limit

Find the limit \( \lim_{x \to 0} \frac{\sqrt{1 + \tan{x}} - \sqrt{1 + \sin{x}}}{x^3} \).

Solution

The limit of the given expression:

lim⁡x→01+tan⁡x−1+sin⁡xx3\lim_{x \to 0} \frac{\sqrt{1 + \tan{x}} - \sqrt{1 + \sin{x}}}{x^3}

is 14\frac{1}{4}.

Would you like more details on how this limit was computed or an explanation of the techniques involved?

Here are some related questions:

  1. How does L'Hopital's rule apply in finding limits?
  2. What is the difference between the behavior of tan⁡x\tan{x} and sin⁡x\sin{x} as x→0x \to 0?
  3. How does the binomial expansion help in simplifying limits with radicals?
  4. What are common techniques to deal with limits involving indeterminate forms?
  5. How does the order of approximation affect small-angle approximations for trigonometric functions?

Tip: For limits involving indeterminate forms like 0/00/0, applying series expansion or using L'Hopital's rule can greatly simplify the calculation.

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
Indeterminate Forms
Series Expansion

Formulas

L'Hopital's Rule: \( \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \) for indeterminate forms
Small angle approximations for \( \tan{x} \approx x \) and \( \sin{x} \approx x \)
Binomial expansion: \( (1 + x)^n \approx 1 + nx \)

Theorems

L'Hopital's Rule
Taylor Series Expansion

Suitable Grade Level

Undergraduate Mathematics or Advanced High School Calculus

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