We are tasked with finding the limit:

$\lim_{x \to 0^+} \left(\cos(x)\right)^{\frac{9}{x^2}}.$

Step 1: Expressing the limit in a more convenient form

We begin by rewriting the expression to use logarithms. If we have an expression of the form $a^b$, we can take the natural logarithm to simplify the process.

Let:

$y = \left(\cos(x)\right)^{\frac{9}{x^2}}.$

Take the natural logarithm of both sides:

$\ln(y) = \ln\left( \left(\cos(x)\right)^{\frac{9}{x^2}} \right) = \frac{9}{x^2} \ln(\cos(x)).$

Now, we need to evaluate the limit of $\ln(y)$ as $x \to 0^+$:

$\lim_{x \to 0^+} \frac{9}{x^2} \ln(\cos(x)).$

Step 2: Approximation of $\ln(\cos(x))$

When $x$ is small, we can use the approximation:

$\cos(x) \approx 1 - \frac{x^2}{2}.$

Thus, for small $x$,

$\ln(\cos(x)) \approx \ln\left( 1 - \frac{x^2}{2} \right).$

Using the approximation $\ln(1 - u) \approx -u$ for small $u$, we have:

$\ln(\cos(x)) \approx -\frac{x^2}{2}.$

Step 3: Substituting the approximation

Now, substitute this approximation into the expression for $\ln(y)$:

$\lim_{x \to 0^+} \frac{9}{x^2} \left( -\frac{x^2}{2} \right) = \lim_{x \to 0^+} \frac{-9x^2}{2x^2} = \lim_{x \to 0^+} \frac{-9}{2} = -\frac{9}{2}.$

Step 4: Exponentiating to find $y$

Recall that we initially defined $y$ as:

$y = \left(\cos(x)\right)^{\frac{9}{x^2}},$

and we found that $\ln(y) \to -\frac{9}{2}$ as $x \to 0^+$. Therefore,

$y \to e^{-\frac{9}{2}}.$

Thus, the limit is:

$\lim_{x \to 0^+} \left(\cos(x)\right)^{\frac{9}{x^2}} = e^{-\frac{9}{2}}.$

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Would you like further explanation on any part of this solution?

Here are five related questions that expand on this topic:

1. What happens if we take the limit of $\left(\sin(x)\right)^{\frac{9}{x^2}}$ as $x \to 0^+$?
2. How can you evaluate limits of the form $\left(1 + f(x)\right)^{g(x)}$ as $x \to 0$?
3. What is the Taylor expansion for $\cos(x)$ around $x = 0$?
4. How do you compute limits involving logarithmic approximations like $\ln(1 + u)$ as $u \to 0$?
5. Can you apply the same method to find the limit of $\left(\tan(x)\right)^{\frac{9}{x^2}}$ as $x \to 0^+$?

Tip: When dealing with limits of exponential functions like this one, logarithmic manipulation often simplifies the problem, especially when the exponent grows unbounded as $x \to 0$.