Calculus
Limits
GRB_1000_SCQ
Grade Class 12

Question:

Let $m$ be a positive integer. If $\lim_{x \to 0} |\cos x + \sin 2x + \sin 3x|^{\cot x} = e^m$, then the value of $m$ is:
2
3
4
5

Step-by-Step Solution

Key Concept: $1^\infty$ form limit using $e^{\lim f(x)\ln g(x)}$
Step 1: Identify the indeterminate form and rewrite the limit using exponentials. We need to evaluate $L = \lim_{x \to 0}|\cos x + \sin 2x + \sin 3x|^{\cot x}$. As $x \to 0$, the base approaches $1$ and the exponent approaches $\infty$, giving us the indeterminate form $1^\infty$. We can rewrite this using the exponential function: $$L = e^{\lim_{x \to 0} \cot x \cdot \ln|\cos x + \sin 2x + \sin 3x|}$$ Step 2: Find the Taylor expansion of the base near $x = 0$. We expand each trigonometric function as $x \to 0$: - $\cos x \approx 1$ - $\sin 2x \approx 2x$ - $\sin 3x \approx 3x$ Therefore: $$\cos x + \sin 2x + \sin 3x \approx 1 + 2x + 3x = 1 + 5x$$ Step 3: Simplify the logarithm of the base. Using the approximation $\ln(1 + u) \approx u$ for small $u$: $$\ln|\cos x + \sin 2x + \sin 3x| = \ln(1 + 5x) \approx 5x$$ Step 4: Find the approximation of $\cot x$ near $x = 0$. As $x \to 0$: $$\cot x = \frac{\cos x}{\sin x} \approx \frac{1}{x}$$ Step 5: Evaluate the exponent in the limit. The exponent becomes: $$\lim_{x \to 0} \cot x \cdot \ln|\cos x + \sin 2x + \sin 3x| = \lim_{x \to 0} \frac{1}{x} \cdot 5x = \lim_{x \to 0} 5 = 5$$ Step 6: State the final answer. Therefore: $$L = e^5$$ Comparing with the given condition $\lim_{x \to 0} |\cos x + \sin 2x + \sin 3x|^{\cot x} = e^m$, we have: $$m = 5$$ The answer is **Option 4: 5**
Correct Answer: 4

Master Calculus with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free