https://file.aiquickdraw.com/mathbot/file/17305748296006uktfp8q.jpg

Let's analyze and solve each of these limits using derivatives, as suggested by the exercise instructions.

### Problem 1:
\[
\lim_{{x \to 1}} \frac{(x+2)^{2018} - 1}{x + 1}
\]

This problem involves the limit of a fraction. To solve it, we can use **L'Hôpital's Rule**, since substituting \( x = 1 \) directly gives the indeterminate form \( \frac{0}{0} \).

Let's apply L'Hôpital's Rule, which requires us to take the derivative of the numerator and the denominator:

1. **Numerator Derivative**: The function in the numerator is \( f(x) = (x+2)^{2018} - 1 \). The derivative \( f'(x) \) is:
   \[
   f'(x) = 2018 \cdot (x+2)^{2017}
   \]

2. **Denominator Derivative**: The denominator is \( g(x) = x + 1 \), and its derivative \( g'(x) \) is:
   \[
   g'(x) = 1
   \]

After taking the derivatives, the limit becomes:
\[
\lim_{{x \to 1}} \frac{2018 \cdot (x+2)^{2017}}{1} = 2018 \cdot (1+2)^{2017} = 2018 \cdot 3^{2017}
\]

So, the solution to the first limit is:
\[
\lim_{{x \to 1}} \frac{(x+2)^{2018} - 1}{x + 1} = 2018 \cdot 3^{2017}
\]

### Problem 2:
\[
\lim_{{x \to \frac{\pi}{6}}} \frac{2 \sin x - 1}{x - \frac{\pi}{6}}
\]

Similar to Problem 1, substituting \( x = \frac{\pi}{6} \) directly in this limit also yields an indeterminate form \( \frac{0}{0} \). We can again apply **L'Hôpital's Rule**:

1. **Numerator Derivative**: The function in the numerator is \( f(x) = 2 \sin x - 1 \). Its derivative is:
   \[
   f'(x) = 2 \cos x
   \]

2. **Denominator Derivative**: The denominator \( g(x) = x - \frac{\pi}{6} \) has a derivative of:
   \[
   g'(x) = 1
   \]

With L'Hôpital's Rule applied, the limit simplifies to:
\[
\lim_{{x \to \frac{\pi}{6}}} \frac{2 \cos x}{1} = 2 \cos \frac{\pi}{6} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
\]

Thus, the solution to the second limit is:
\[
\lim_{{x \to \frac{\pi}{6}}} \frac{2 \sin x - 1}{x - \frac{\pi}{6}} = \sqrt{3}
\]

### Summary of Answers:
1. \(\lim_{{x \to 1}} \frac{(x+2)^{2018} - 1}{x + 1} = 2018 \cdot 3^{2017}\)
2. \(\lim_{{x \to \frac{\pi}{6}}} \frac{2 \sin x - 1}{x - \frac{\pi}{6}} = \sqrt{3}\)

Would you like a more detailed explanation of each step, or do you have any questions?

#### Related Questions:
1. What are the conditions for applying L'Hôpital's Rule?
2. How would the solutions change if the exponents in Problem 1 were different?
3. Can you explain why substitution gives an indeterminate form in these problems?
4. Are there alternative methods to solve these limits without L'Hôpital's Rule?
5. How does the derivative of trigonometric functions affect limits?

#### Tip:
When approaching limits that yield \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), consider L'Hôpital's Rule, but also explore factoring, simplification, or series expansion, as these might sometimes offer simpler solutions.

Using derivatives to calculate the following limits:

1) \( \lim_{{x \to 1}} \frac{(x+2)^{2018} - 1}{x + 1} \)

2) \( \lim_{{x \to \frac{\pi}{6}}} \frac{2 \sin x - 1}{x - \frac{\pi}{6}} \)

Learn how to calculate complex limits involving derivatives and trigonometric functions. This example uses L'Hôpital's Rule to solve limits with indeterminate forms. Ideal for undergraduate Calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation