Thank you for providing the image. I'll go through each of the limit problems and provide a solution. Let's analyze them one by one.

Problem 3:

$\lim_{x \to +\infty} \left(\frac{1 + 1/x}{x}\right)$

Solution for Problem 3:

As $x \to +\infty$, $\frac{1}{x} \to 0$. Therefore, $1 + \frac{1}{x} \to 1$. $\frac{1 + \frac{1}{x}}{x} \approx \frac{1}{x} \to 0 \text{ as } x \to +\infty.$ Answer: 0.

Problem 4:

$\lim_{x \to 0^-} \frac{\sin x + \tan x}{x}$

Solution for Problem 4:

As $x \to 0^-$, both $\sin x$ and $\tan x$ can be approximated by $x$, thus: $\frac{\sin x + \tan x}{x} \approx \frac{x + x}{x} = 2.$ Answer: 2.

Problem 5:

$\lim_{x \to +\infty} \frac{1}{\sqrt{x^2 + x} - x}$

Solution for Problem 5:

Rationalize the denominator: $\frac{1}{\sqrt{x^2 + x} - x} \cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x} = \frac{\sqrt{x^2 + x} + x}{x}.$ Then, as $x \to +\infty$, this simplifies to 1. Answer: 1.

Problem 6:

$\lim_{x \to 0^+} \frac{1}{x^{1/2}}$

Solution for Problem 6:

As $x \to 0^+$, $x^{1/2} \to 0$, making $\frac{1}{x^{1/2}}$ diverge to $+\infty$. Answer: $+\infty$.

Problem 7:

$\lim_{x \to 0} ((\sin x)^2 + (\cos x)^2)$

Solution for Problem 7:

Using the Pythagorean identity: $(\sin x)^2 + (\cos x)^2 = 1.$ Answer: 1.

Problem 8:

$\lim_{x \to +\infty} \left(\frac{1 + 1/x}{x} - 2x\right)$

Solution for Problem 8:

First term $\to 0$