Question

\( \lim_{x \to 0} \frac{\sqrt{\cos 2x + 3} - \sqrt{\cos^2 x + \sin x + 3}}{x} \) is equal to

• A. \( \frac{1}{4} \)
• B. \( -\frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( -\frac{1}{2} \)
• E. \( -1 \)

Hint:

For limits of type \( \frac{f(x)-f(0)}{x} \), directly use derivative at 0.

Answer 1

The Correct Option is B

Note: The OCR'd question seems to have a typo (`cos2x` vs `cos^2x`). Based on the image, the limit is \(\lim_{x \to 0} \frac{\sqrt{\cos^2 x + 3} - \sqrt{\cos^2 x + \sin x + 3}}{x}\).
Step 1: Understanding the Concept:
This is a limit of the form \(\frac{0}{0}\), which can be solved by multiplying by the conjugate, or by recognizing the structure of a derivative.
Step 2: Key Formula or Approach:
Method 1: Multiplying by the Conjugate. Multiply the numerator and denominator by \((\sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3})\). Method 2: Using the definition of a derivative. The limit is in the form \(\lim_{h \to 0} \frac{f(c+h)-f(c)}{h}\), which is the definition of \(f'(c)\). Here, we can define a function and evaluate its derivative.
Step 3: Detailed Explanation:
Method 1: Multiplying by the Conjugate \[ \lim_{x \to 0} \frac{\sqrt{\cos^2 x + 3} - \sqrt{\cos^2 x + \sin x + 3}}{x} \times \frac{\sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3}}{\sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3}} \] The numerator becomes: \[ (\cos^2 x + 3) - (\cos^2 x + \sin x + 3) = \cos^2 x + 3 - \cos^2 x - \sin x - 3 = -\sin x \] The expression is now: \[ \lim_{x \to 0} \frac{-\sin x}{x(\sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3})} \] We can split this into two parts: \[ \lim_{x \to 0} \left(-\frac{\sin x}{x}\right) \cdot \lim_{x \to 0} \frac{1}{\sqrt{\cos^2 x + 3} + \sqrt{\cos^2 x + \sin x + 3}} \] The first limit is a standard one: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\). For the second limit, we can substitute \(x=0\): \[ \frac{1}{\sqrt{\cos^2(0) + 3} + \sqrt{\cos^2(0) + \sin(0) + 3}} = \frac{1}{\sqrt{1+3} + \sqrt{1+0+3}} = \frac{1}{\sqrt{4} + \sqrt{4}} = \frac{1}{2+2} = \frac{1}{4} \] Combining the results: \[ -1 \cdot \frac{1}{4} = -\frac{1}{4} \] Method 2: Using Derivatives Let \(f(t) = \sqrt{t+3}\). Let \(g(x) = \cos^2 x\) and \(h(x) = \cos^2 x + \sin x\). The limit can be written as \(\lim_{x \to 0} \frac{f(g(x)) - f(h(x))}{x}\). This isn't a direct derivative form. A better approach: Let \(F(x) = \sqrt{\cos^2 x + \sin x + 3}\). Then \(F(0) = \sqrt{1+0+3}=2\). Let \(G(x) = \sqrt{\cos^2 x + 3}\). Then \(G(0) = \sqrt{1+3}=2\). The limit is \(\lim_{x \to 0} \frac{G(x) - F(x)}{x} = \lim_{x \to 0} \frac{(G(x)-G(0)) - (F(x)-F(0))}{x} = G'(0) - F'(0)\). \(G'(x) = \frac{-2\cos x \sin x}{2\sqrt{\cos^2 x+3}}\), so \(G'(0)=0\). \(F'(x) = \frac{-2\cos x \sin x + \cos x}{2\sqrt{\cos^2 x+\sin x+3}}\), so \(F'(0) = \frac{1}{2\sqrt{4}} = \frac{1}{4}\). The limit is \(G'(0) - F'(0) = 0 - \frac{1}{4} = -\frac{1}{4}\). Step 4: Final Answer:
The limit is \(-\frac{1}{4}\).