Math Problem Statement

lim_{x \to 0} \frac{\sec x - 1}{\tan^2 x}

Solution

The limit of the given expression limx0secx1tan2x\lim_{x \to 0} \frac{\sec x - 1}{\tan^2 x} is 12\frac{1}{2}.

Would you like a detailed explanation on how this limit was solved?

Here are 5 related questions to explore:

  1. What is the relationship between sec(x)\sec(x) and tan(x)\tan(x)?
  2. How can we simplify limits involving trigonometric functions near zero?
  3. What is L'Hopital's rule and when can it be applied to limits?
  4. How can the small angle approximations for sine and cosine help in solving limits?
  5. How can you find the limit of a rational trigonometric function using series expansion?

Tip: When dealing with trigonometric limits near zero, small-angle approximations like tanxx\tan x \approx x and secx1+x22\sec x \approx 1 + \frac{x^2}{2} are very useful.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Limits

Formulas

Small angle approximations: \sec x \approx 1 + \frac{x^2}{2}, \tan x \approx x
L'Hopital's Rule

Theorems

L'Hopital's Rule
Trigonometric limit properties

Suitable Grade Level

Grades 11-12 (Pre-Calculus and Calculus)