Question

Find \( \alpha \) if \[ \lim_{x \to 0} \frac{1 - \sec^2(\alpha x)}{\alpha x^2} = -3 \]

Hint:

When solving limits with trigonometric functions, use approximations like \( \tan(x) \approx x \) for small angles.

Answer 1

Step 1: Simplify the given expression.
We are given the limit: \[ \lim_{x \to 0} \frac{1 - \sec^2(\alpha x)}{\alpha x^2} = -3 \] To solve this, first use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \). Hence, \[ \sec^2(\alpha x) = 1 + \tan^2(\alpha x) \] Substitute this into the limit expression: \[ \lim_{x \to 0} \frac{1 - (1 + \tan^2(\alpha x))}{\alpha x^2} = -3 \] This simplifies to: \[ \lim_{x \to 0} \frac{- \tan^2(\alpha x)}{\alpha x^2} = -3 \]
Step 2: Use the approximation for small angles.
For small angles, \( \tan(\theta) \approx \theta \), so: \[ \tan(\alpha x) \approx \alpha x \] Thus, \[ \tan^2(\alpha x) \approx (\alpha x)^2 \] Substitute this approximation into the expression: \[ \lim_{x \to 0} \frac{- (\alpha x)^2}{\alpha x^2} = -3 \] Simplify the expression: \[ \lim_{x \to 0} -\alpha^2 = -3 \]
Step 3: Solve for \( \alpha \).
This gives: \[ \alpha^2 = 3 \] Hence, \[ \alpha = \pm \sqrt{3} \]