Question

$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}$ is equal to:

• A. 2
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The Correct Option is C

1. Understand the problem:

We need to evaluate the limit:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} \]

2. Substitute \( x = \frac{\pi}{4} \) directly:

First, check if direct substitution leads to an indeterminate form:

\[ \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}, \quad \cot \left( \frac{\pi}{4} \right) = 1 \]

Substituting these values:

\[ \frac{\sqrt{2} \cdot \frac{\sqrt{2}}{2} - 1}{1 - 1} = \frac{1 - 1}{0} = \frac{0}{0} \]

This is an indeterminate form, so we need to apply L'Hôpital's Rule.

3. Apply L'Hôpital's Rule:

Differentiate the numerator and denominator with respect to \( x \):

Numerator derivative:

\[ \frac{d}{dx} \left( \sqrt{2} \cos x - 1 \right) = -\sqrt{2} \sin x \]

Denominator derivative:

\[ \frac{d}{dx} \left( \cot x - 1 \right) = -\csc^2 x \]

Now, the limit becomes:

\[ \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\csc^2 x} = \lim_{x \to \frac{\pi}{4}} \sqrt{2} \sin x \sin^2 x = \sqrt{2} \sin^3 x \]

Substitute \( x = \frac{\pi}{4} \):

\[ \sqrt{2} \left( \frac{\sqrt{2}}{2} \right)^3 = \sqrt{2} \cdot \frac{2 \sqrt{2}}{8} = \frac{4}{8} = \frac{1}{2} \]

Correct Answer: (C) \(\frac{1}{2}\)

Answer 2

We begin by substituting $x = \frac{\pi}{4}$: \[ \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}. \] For small values around $x = \frac{\pi}{4}$, use the approximations: \[ \cos x \approx 1 - \frac{(x - \frac{\pi}{4})^2}{2} \quad \text{and} \quad \cot x \approx 1 + (x - \frac{\pi}{4}). \] Substituting these approximations and simplifying, we find the limit to be $\frac{1}{2}$.