If \( \alpha = \lim_{x \to \pi/4} \frac{\tan^3 x - \tan x}{\cos(x + \frac{\pi}{4})} \) and \( \beta = \lim_{x \to 0} (\cos x)^{\cot x} \) are the roots of the equation, \( ax^2 + bx - 4 = 0 \), then the ordered pair \( (a, b) \) is :
Show Hint
Limits of the form \( (\cos x)^{\cot x} \) as \( x \to 0 \) involve the \( 1^\infty \) indeterminate form.
Recall that \( \cos x \approx 1 - \frac{x^2}{2} \) and \( \tan x \approx x \), so the exponent limit is \( \frac{-x^2/2}{x} \to 0 \).
Step 1: Understanding the Concept:
We need to evaluate two limits to find the roots \( \alpha \) and \( \beta \).
Then, we use the sum and product of roots to determine the coefficients \( a \) and \( b \). Step 2: Key Formula or Approach:
For \( \alpha \), use L'Hopital's rule or trigonometric expansion.
For \( \beta \), use the \( 1^\infty \) form formula: \( \lim f(x)^{g(x)} = e^{\lim g(x)(f(x)-1)} \). Step 3: Detailed Explanation:
Evaluate \( \alpha \):
\[ \alpha = \lim_{x \to \pi/4} \frac{\tan x (\tan^2 x - 1)}{\cos x \frac{1}{\sqrt{2}} - \sin x \frac{1}{\sqrt{2}}} = \sqrt{2} \lim_{x \to \pi/4} \frac{\tan x (\tan^2 x - 1)}{\cos x - \sin x} \]
Applying L'Hopital's Rule:
Numerator derivative at \( \pi/4 \): \( [\sec^2 x (\tan^2 x - 1) + \tan x (2 \tan x \sec^2 x)]_{x=\pi/4} = 0 + 1(2 \cdot 1 \cdot 2) = 4 \).
Denominator derivative at \( \pi/4 \): \( [-\sin x - \cos x]_{x=\pi/4} = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} \).
So, \( \alpha = \frac{4}{-\sqrt{2}} \cdot \sqrt{2} = -4 \).
Evaluate \( \beta \):
\[ \beta = \lim_{x \to 0} (\cos x)^{\cot x} = e^{\lim_{x \to 0} \cot x (\cos x - 1)} = e^{\lim_{x \to 0} \frac{\cos x - 1}{\tan x}} \]
Using L'Hopital: \( \lim_{x \to 0} \frac{-\sin x}{\sec^2 x} = 0 \).
So, \( \beta = e^0 = 1 \).
The roots are \( -4 \) and \( 1 \).
Equation: \( (x + 4)(x - 1) = 0 \implies x^2 + 3x - 4 = 0 \).
Comparing with \( ax^2 + bx - 4 = 0 \), we get \( a = 1, b = 3 \). Step 4: Final Answer:
The ordered pair \( (a, b) \) is \( (1, 3) \).
