The given limit is:

$\lim_{{x \to 0}} \left(1 + \sin x\right)^{\cot x}$

To evaluate this limit, we can use the exponential form to handle the indeterminate expression. Let's rewrite it as:

$\lim_{{x \to 0}} \left(1 + \sin x\right)^{\cot x} = e^{\lim_{{x \to 0}} \cot x \cdot \ln(1 + \sin x)}$

Now, we will evaluate the inner limit:

$\lim_{{x \to 0}} \cot x \cdot \ln(1 + \sin x)$

We know that $\cot x = \frac{\cos x}{\sin x}$ and, as $x \to 0$, $\sin x \approx x$ and $\cos x \approx 1$, so $\cot x \approx \frac{1}{x}$. Also, for small values of $x$, $\ln(1 + \sin x) \approx \sin x = x$.

Thus, the limit simplifies to:

$\lim_{{x \to 0}} \frac{\ln(1 + \sin x)}{\sin x} \approx \lim_{{x \to 0}} \frac{x}{x} = 1$

Therefore, the exponential form of the limit is:

$e^{\lim_{{x \to 0}} \cot x \cdot \ln(1 + \sin x)} = e^{1} = e$

The final answer is:

$\lim_{{x \to 0}} \left(1 + \sin x\right)^{\cot x} = e$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would you evaluate the limit of $(1 + \sin x)^{1/x}$ as $x \to 0$?
2. What is the general approach to solving limits involving exponential forms?
3. How does L'Hospital's Rule apply to limits like this?
4. How do you determine when a limit tends towards $e$ in indeterminate forms?
5. What are the conditions under which $\lim_{{x \to 0}} \left(1 + x\right)^{\frac{1}{x}} = e$?

Tip: For limits involving forms like $0^0$, $1^\infty$, or $\infty^0$, rewriting the expression in exponential form is a helpful strategy.