I have displayed the image. Now, I will extract and solve the mathematical problems from questions 15 to 20.

Question 15

The problem asks to find the value of: $\lim_{x \to \infty} \left( \frac{\sin x}{x} - \sin 7x \right)$ Solution:
Since $\sin x$ is bounded between -1 and 1, as $x \to \infty$, we have: $\frac{\sin x}{x} \to 0 \quad \text{and} \quad \sin 7x \text{ oscillates between -1 and 1.}$ Thus, the limit is dominated by the $\frac{\sin x}{x}$ term going to zero, and the overall limit is 0.
Answer: 0.

Question 16

Find the limit: $\lim_{x \to 0} \frac{1 - \sin^2 x}{x}$ Solution: We rewrite $1 - \sin^2 x$ as $\cos^2 x$, so the expression becomes: $\lim_{x \to 0} \frac{\cos^2 x}{x}$ At $x = 0$, this limit approaches infinity, as the numerator is close to 1, while the denominator approaches 0.
Answer: $\infty$.

Question 17

The limit to evaluate is: $\lim_{x \to 0} \frac{\sin(2x) - 6x}{x^3}$ Solution: Using Taylor expansions for small $x$, we get: $\sin(2x) \approx 2x - \frac{(2x)^3}{6} = 2x - \frac{8x^3}{6}$ Thus, the expression becomes: $\lim_{x \to 0} \frac{(2x - \frac{8x^3}{6}) - 6x}{x^3} = \lim_{x \to 0} \frac{-4x}{x^3} = -\lim_{x \to 0} \frac{4}{x^2} = -\infty$ Answer: $-\infty$.

Question 18

Evaluate: $\lim_{x \to \infty} \frac{\ln(2x) - 6}{x}$ Solution: We can split the limit as follows: $\lim_{x \to \infty} \frac{\ln(2x)}{x} - \lim_{x \to \infty} \frac{6}{x}$ The first term goes to 0 because $\ln(2x)$ grows slower than $x$, and the second term also goes to 0. Therefore, the limit is 0.
Answer: 0.

Question 19

The limit to solve is: $\lim_{x \to 0} \frac{\tan(2x) - 6x}{x^3}$ Solution: Using the Taylor series for $\tan(2x)$, we approximate $\tan(2x)$ for small $x$: $\tan(2x) \approx 2x + \frac{(2x)^3}{3}$ Thus, the expression becomes: $\lim_{x \to 0} \frac{(2x + \frac{8x^3}{3}) - 6x}{x^3} = \lim_{x \to 0} \frac{-4x + \frac{8x^3}{3}}{x^3} = \lim_{x \to 0} \frac{-4}{x^2} = -\infty$ Answer: $-\infty$.

Question 20

Evaluate the following limit: $\lim_{x \to 1} \frac{\sqrt{x^3 + \cos(2x)} - 2}{x - 1}$ Solution: We use L'Hopital's Rule because we have an indeterminate form $\frac{0}{0}$. Differentiating the numerator and denominator:

• Derivative of the numerator: $\frac{d}{dx} \left( \sqrt{x^3 + \cos(2x)} \right) = \frac{3x^2 - 2\sin(2x)}{2\sqrt{x^3 + \cos(2x)}}$
• Derivative of the denominator: 1

Substitute $x = 1$: $\frac{3(1)^2 - 2\sin(2)}{2\sqrt{1^3 + \cos(2)}} = \frac{3 - 2\sin(2)}{2\sqrt{1 + \cos(2)}}$ This gives the final value after evaluation (needs a calculator for precise value).
Answer: Exact numeric value.

Would you like further explanations on any specific problem? Here are some related questions for practice:

1. Can you expand the Taylor series for other trigonometric functions?
2. How does L'Hopital's rule apply to indeterminate forms?
3. Why do trigonometric functions like $\sin(x)$ oscillate but still have limits at infinity?
4. How can limits involving logarithmic functions be simplified?
5. What is the geometric interpretation of limits involving square roots?

Tip: When evaluating complex limits, always check for indeterminate forms to decide if tools like L'Hopital's rule or series expansions are needed.