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Let's go over the limits one by one:

### 1. \(\lim_{{x \to 0}} \frac{{\sin(2x)}}{{\sin(5x)}}\)

Using the small angle approximation \(\sin x \approx x\) for \(x \to 0\):

\[
\lim_{{x \to 0}} \frac{{\sin(2x)}}{{\sin(5x)}} = \lim_{{x \to 0}} \frac{{2x}}{{5x}} = \frac{2}{5}.
\]

### 2. \(\lim_{{x \to 0}} x \cot x\)

We know that \(\cot x = \frac{{\cos x}}{{\sin x}}\). Therefore:

\[
x \cot x = x \cdot \frac{{\cos x}}{{\sin x}} = \frac{{x \cos x}}{{\sin x}}.
\]

As \(x \to 0\), \(\cos x \to 1\) and \(\sin x \approx x\), so:

\[
\lim_{{x \to 0}} \frac{{x \cos x}}{{\sin x}} = \lim_{{x \to 0}} \frac{{x}}{{x}} = 1.
\]

### 3. \(\lim_{{x \to 0}} \frac{{\arctan x}}{{x}}\)

Using the fact that \(\lim_{{x \to 0}} \frac{{\arctan x}}{{x}} = 1\), we directly obtain:

\[
\lim_{{x \to 0}} \frac{{\arctan x}}{{x}} = 1.
\]

### 4. \(\lim_{{x \to \infty}} \left(1 + \frac{1}{x}\right)^{\frac{x}{2}}\)

This is a standard limit, which can be rewritten in the form of an exponential function. We recognize that:

\[
\lim_{{x \to \infty}} \left(1 + \frac{1}{x}\right)^{x} = e,
\]

so:

\[
\lim_{{x \to \infty}} \left(1 + \frac{1}{x}\right)^{\frac{x}{2}} = e^{1/2} = \sqrt{e}.
\]

### 5. \(\lim_{{x \to \infty}} \left(\frac{{x+3}}{{x-2}}\right)^{2x+1}\)

Simplifying the fraction:

\[
\frac{{x+3}}{{x-2}} = \frac{{1 + \frac{3}{x}}}{{1 - \frac{2}{x}}} \to 1 \quad \text{as} \quad x \to \infty.
\]

Thus, the limit becomes:

\[
\lim_{{x \to \infty}} \left(1\right)^{2x+1} = 1.
\]

### 6. \(\lim_{{x \to 0}} \left(1 + 3 \tan^2 x\right)^{\cot^2 x}\)

As \(x \to 0\), \(\tan x \approx x\) and \(\cot x \approx \frac{1}{x}\). Therefore, we have:

\[
\left(1 + 3 \tan^2 x\right) \approx 1 + 3x^2.
\]

Now,

\[
\cot^2 x \approx \frac{1}{x^2},
\]

so the limit becomes:

\[
\lim_{{x \to 0}} \left(1 + 3x^2\right)^{\frac{1}{x^2}}.
\]

This is an indeterminate form of the type \(1^\infty\), so we take the logarithm:

\[
\lim_{{x \to 0}} \frac{\ln(1 + 3x^2)}{x^2} = 3.
\]

Thus, the limit is:

\[
e^3.
\]

Would you like any specific details or further explanation on these? Here are some related questions for further practice:

1. What is the limit \(\lim_{{x \to 0}} \frac{\sin 3x}{\sin 4x}\)?
2. How would you find \(\lim_{{x \to \infty}} \left(1 - \frac{1}{x}\right)^x\)?
3. What happens if you replace \(\arctan x\) with \(\tan^{-1}(x^2)\) in problem 3?
4. How does L'Hopital's Rule apply to the form \(\frac{0}{0}\) in problem 1?
5. How would you calculate \(\lim_{{x \to \infty}} \left(1 + \frac{2}{x}\right)^{x/2}\)?

**Tip:** For limits involving trigonometric functions near \(x = 0\), always consider using approximations like \(\sin x \approx x\) or \(\tan x \approx x\).

求下列极限: 
(1) lim_{x → 0} sin(2x)/sin(5x);
(2) lim_{x → 0} x cot(x);
(3) lim_{x → 0} arctan(x)/x;
(4) lim_{x → ∞} (1 + 1/x)^(x/2);
(5) lim_{x → ∞} ((x + 3)/(x - 2))^(2x + 1);
(6) lim_{x → 0} (1 + 3 tan^2(x))^(cot^2(x))

Detailed step-by-step solutions for limit problems involving trigonometric and indeterminate forms. These include solving lim_{x → 0} sin(2x)/sin(5x), lim_{x → ∞} (1 + 1/x)^(x/2), and more. Useful for university students studying calculus.

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