Question

Let \(k\in R\). If \(\)\(\)\(\)\(\lim\limits_{x→0+}(\sin(\sin kx)+\cos x+x)^{\frac{2}{x}}=e^6\), then the value of k is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

1. Understanding the Problem:
We are given the following expression:

$ I = \lim_{x \to 0^+} \left( \sin(kx) + \cos(x + x) \right)^2 x = e^6 $

2. Taking the Natural Logarithm:
We begin by taking the natural logarithm of both sides of the equation:

$ \ln I = \lim_{x \to 0^+} \frac{2}{x} \left( \sin(\sin(kx)) + \cos(x + x) - 1 \right) $

3. Simplifying the Expression:
Now we simplify the expression by analyzing the terms:

$ \ln I = \lim_{x \to 0^+} 2 \left( \frac{\sin(\sin(kx)) \cdot \sin(kx) \cdot kx}{\sin(kx)} + \frac{1 - \cos(x)}{x^2} \right) $

4. Evaluating the Limit:
Taking the limit as $ x \to 0^+ $ and applying known limits for sine and cosine functions, we get:

$ \ln I = 2(k + 1) $

5. Solving for $k$:
Since we are given that $ I = e^{6} $, equating both sides results in:

$ 2(k + 1) = 6 $

6. Final Calculation:
Solving for $ k $:

$ k + 1 = 3 \Rightarrow k = 2 $

Final Answer:
Thus, the value of $ k $ is $ 2 $.

Answer 2

To solve the problem, we need to find the value of \(k \in \mathbb{R}\) such that:

\[ \lim_{x \to 0^+} \left(\sin(\sin kx) + \cos x + x \right)^{\frac{2}{x}} = e^6 \]

1. Simplify the expression inside the limit:
As \(x \to 0\), use the expansions:

\[ \sin(\sin kx) \approx \sin(kx) \approx kx \] (since \(\sin \theta \approx \theta\) for small \(\theta\)), \[ \cos x \approx 1 - \frac{x^2}{2} \]

So inside the parentheses, approximate:

\[ \sin(\sin kx) + \cos x + x \approx kx + \left(1 - \frac{x^2}{2}\right) + x = 1 + (k + 1)x - \frac{x^2}{2} \]

2. Rewrite the limit in exponential form:
Let \[ L = \lim_{x \to 0^+} \left(1 + (k+1)x - \frac{x^2}{2}\right)^{\frac{2}{x}} \] Take natural log: \[ \ln L = \lim_{x \to 0^+} \frac{2}{x} \ln \left(1 + (k+1)x - \frac{x^2}{2}\right) \] Using \(\ln(1 + y) \approx y\) for small \(y\), we get: \[ \ln L = \lim_{x \to 0^+} \frac{2}{x} \left((k+1)x - \frac{x^2}{2}\right) = \lim_{x \to 0^+} 2(k+1) - x = 2(k+1) \]

3. Given:
\[ L = e^6 \Rightarrow \ln L = 6 \] So, \[ 2(k+1) = 6 \implies k + 1 = 3 \implies k = 2 \]

Final Answer:
\[ \boxed{2} \]