We aim to evaluate the limit:

$\lim_{x \to 0} \left( 2\cos(x)^{3/2} - \frac{\sin(x)}{\arctan(x)} \right)^{\frac{x}{\tan(x) - \sinh(x)}}.$

Step 1: Analyze the Base

The base of the exponential function is:

$f(x) = 2\cos(x)^{3/2} - \frac{\sin(x)}{\arctan(x)}.$

First Term: $2\cos(x)^{3/2}$

As $x \to 0$, $\cos(x) \to 1$. Thus: $2\cos(x)^{3/2} \to 2.$

Second Term: $-\frac{\sin(x)}{\arctan(x)}$

As $x \to 0$, both $\sin(x)$ and $\arctan(x)$ behave like $x$, so: $\frac{\sin(x)}{\arctan(x)} \to 1.$

Thus, the base $f(x)$ simplifies to: $f(x) \to 2 - 1 = 1 \quad \text{as } x \to 0.$

Step 2: Analyze the Exponent

The exponent of the function is:

$g(x) = \frac{x}{\tan(x) - \sinh(x)}.$

Expand $\tan(x)$ and $\sinh(x)$:

Using their Taylor series expansions around $x = 0$: $\tan(x) \sim x + \frac{x^3}{3}, \quad \sinh(x) \sim x + \frac{x^3}{6}.$

Thus: $\tan(x) - \sinh(x) \sim \left(x + \frac{x^3}{3}\right) - \left(x + \frac{x^3}{6}\right) = \frac{x^3}{6}.$

So, the exponent becomes: $g(x) = \frac{x}{\frac{x^3}{6}} = \frac{6}{x^2}.$

Step 3: Combine Results

The overall limit is now expressed as: $\lim_{x \to 0} f(x)^{g(x)} = \lim_{x \to 0} \left( 1 \right)^{\frac{6}{x^2}}.$

Since $f(x) \to 1$, we can analyze the behavior more carefully:

1. If $g(x) \to \infty$, $1^{\infty}$ tends to 1 if $f(x)$ is sufficiently close to 1.
2. Here, $f(x) - 1 \to 0$ fast enough that the limit evaluates to: $\boxed{1}.$

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Do you want details about any step or further explanation? 😊

Relative Questions:

1. How can you formally prove the Taylor expansion used in this example?
2. What are alternative approaches to handling $1^{\infty}$ forms in limits?
3. Why does $f(x)$ approaching 1 matter in this context?
4. How is the behavior of $\tan(x)$ and $\sinh(x)$ crucial to determining the exponent?
5. What happens if $f(x) \neq 1$ in such limits?

Tip:

Always simplify expressions using Taylor series when analyzing limits around small values!