$$\lim_{x \to 0} \left(\frac{(x+2\cos x)^3 + 2(x+2\cos x)^2 + 3\sin(x+2\cos x)}{(x+2)^3 + 2(x+2)^2 + 3\sin(x+2)}\right)^{\frac{100}{x}}$$ is equal to

We need to evaluate $$\lim_{x \to 0} \left(\frac{(x+2\cos x)^3 + 2(x+2\cos x)^2 + 3\sin(x+2\cos x)}{(x+2)^3 + 2(x+2)^2 + 3\sin(x+2)}\right)^{\frac{100}{x}}$$.

Let $$g(u) = u^3 + 2u^2 + 3\sin u$$.

The expression becomes $$\left(\frac{g(x + 2\cos x)}{g(x + 2)}\right)^{100/x}$$.

At $$x = 0$$: $$x + 2\cos x = 0 + 2(1) = 2$$ and $$x + 2 = 2$$.

So the base $$\to \frac{g(2)}{g(2)} = 1$$ and the exponent $$\to \infty$$. This is a $$1^\infty$$ indeterminate form.

For a limit of the form $$\lim(1 + h)^{1/\epsilon}$$ where $$h \to 0$$ and $$\epsilon \to 0$$, we use:

$$L = e^{\lim h/\epsilon}$$

Let $$R = \frac{g(x+2\cos x)}{g(x+2)}$$. Then $$R - 1 = \frac{g(x+2\cos x) - g(x+2)}{g(x+2)}$$.

We need $$\ln L = \lim_{x \to 0} \frac{100}{x} \cdot (R - 1)$$ (since $$\ln(1+t) \approx t$$ when $$t \to 0$$, and $$R - 1 \to 0$$).

Let $$u = x + 2$$ and $$v = x + 2\cos x$$.

$$v - u = 2\cos x - 2$$

Using the Taylor expansion $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$:

$$v - u = 2\left(1 - \frac{x^2}{2} + \cdots\right) - 2 = -x^2 + O(x^4)$$

By the Mean Value Theorem / first-order Taylor expansion: when $$v - u$$ is small,

$$g(v) - g(u) \approx g'(u) \cdot (v - u)$$

This approximation is valid because $$g$$ is smooth and $$v - u \to 0$$ as $$x \to 0$$.

$$g'(u) = 3u^2 + 4u + 3\cos u$$

$$g'(2) = 12 + 8 + 3\cos 2 = 20 + 3\cos 2$$

$$g(2) = 8 + 8 + 3\sin 2 = 16 + 3\sin 2$$

$$R - 1 \approx \frac{g'(2) \cdot (-x^2)}{g(2)} = \frac{-(20 + 3\cos 2)}{16 + 3\sin 2} \cdot x^2$$

$$\ln L = \lim_{x \to 0} \frac{100}{x} \cdot \frac{-(20 + 3\cos 2)}{16 + 3\sin 2} \cdot x^2 = \lim_{x \to 0} \frac{-100(20 + 3\cos 2)}{16 + 3\sin 2} \cdot x = 0$$

Since the expression is proportional to $$x$$ which goes to $$0$$, we get $$\ln L = 0$$.

$$L = e^0 = 1$$

The answer is 1.