Question:

Step 1: Analyze the limit structure. As $x \to 0$, we have: $\cos x \to 1$ $\sin 2x \to 0$ $\sin 3x \to 0$ Thus, the base $|\cos x + \sin 2x + \sin 3x| \to |1+0+0| = 1$. Also, $\cot x = \frac{\cos x}{\sin x} \to \infty$ as $x \to 0^+$. This is the indeterminate form $1^\infty$. Step 2: Apply the limit formula for $1^\infty$ forms. For a limit of the form $\displaystyle\lim_{x \to a} [f(x)]^{g(x)}$ where $f(x) \to 1$ and $g(x) \to \infty$, the limit is $e^L$, where $L = \displaystyle\lim_{x \to a} g(x) \cdot (f(x)-1)$. Step 3: Expand the functions using Taylor series around $x=0$. $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$ $\sin 2x = 2x - \frac{(2x)^3}{3!} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5)$ $\sin 3x = 3x - \frac{(3x)^3}{3!} + O(x^5) = 3x - \frac{9x^3}{2} + O(x^5)$ Step 4: Sum the expansions to find the base function $f(x) = \cos x + \sin 2x + \sin 3x$. $f(x) = \left(1 - \frac{x^2}{2} + O(x^4)\right) + \left(2x - \frac{4x^3}{3} + O(x^5)\right) + \left(3x - \frac{9x^3}{2} + O(x^5)\right)$ $f(x) = 1 + (2x+3x) - \frac{x^2}{2} - \left(\frac{4x^3}{3} + \frac{9x^3}{2}\right) + O(x^4)$ $f(x) = 1 + 5x - \frac{x^2}{2} - \left(\frac{8x^3+27x^3}{6}\right) + O(x^4)$ $f(x) = 1 + 5x - \frac{x^2}{2} - \frac{35x^3}{6} + O(x^4)$ Step 5: Determine $f(x)-1$. $f(x)-1 = 5x - \frac{x^2}{2} - \frac{35x^3}{6} + O(x^4)$ Step 6: Compute the limit $L = \displaystyle\lim_{x \to 0} \cot x \cdot (f(x)-1)$. The Taylor expansion for $\cot x$ around $x=0$ is: $\cot x = \frac{1}{x} - \frac{x}{3} + O(x^3)$ Now, we compute the product: $L = \lim_{x \to 0} \left( \frac{1}{x} - \frac{x}{3} + O(x^3) \right) \left( 5x - \frac{x^2}{2} - \frac{35x^3}{6} + O(x^4) \right)$ To find the limit as $x \to 0$, we only need the constant term in the product of the series expansions. The constant term arises from multiplying the lowest order terms: $\left(\frac{1}{x}\right) \cdot (5x) = 5$ All other products will contain powers of $x$ that tend to zero as $x \to 0$: $\left(\frac{1}{x}\right) \cdot \left(-\frac{x^2}{2}\right) = -\frac{x}{2} \to 0$ $\left(-\frac{x}{3}\right) \cdot (5x) = -\frac{5x^2}{3} \to 0$ Therefore, $L = 5$. Step 7: Equate the result to $e^m$. The limit is $e^L = e^5$. Given that $\displaystyle\lim_{x \to 0} |\cos x + \sin 2x + \sin 3x|^{\cot x} = e^m$, we have: $e^m = e^5$ Thus, $m=5$. Since $m$ must be a positive integer, $m=5$ is consistent. The final answer is $\boxed{5}$.