Question

Use Maclaurin series to do and check your results by computer. $$\lim _{x \rightarrow 0}\left(\frac{1+x}{x}-\frac{1}{\sin x}\right)$$

Short answer

The limit is 1.

Step-by-step solution

- Write the Maclaurin series

Start by writing the Maclaurin series expansion for \ \ \(1+x\) and \(\sin x\). \ \ \(1+x = 1 + x\) \ \ For \(\sin x\), we have: \ \ \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\)

- Expand the expression

Using the Maclaurin series, rewrite the expression \ \(\frac{1+x}{x} - \frac{1}{\sin x}\). \ \ Substitute the series for \(\sin x\): \ \ \(\frac{1 + x}{x} - \frac{1}{x - \frac{x^3}{6} + \ldots}\)

- Simplify the denominator

For small \(x\), the higher order terms can be neglected. Thus, the expression becomes: \ \ \(\frac{1 + x}{x} - \frac{1}{x}\)

- Combine the fractions

Combine the fractions to create a common denominator: \ \ \(\frac{1 + x - 1}{x}\) = \(\frac{x}{x}\)

- Simplify the expression

Simplify the combined expression: \ \ \(\frac{x}{x} =1\).

- Determine the limit

Therefore, the limit as \(x \rightarrow 0\) is: \ \ \ \lim_{x \to 0} \left(\frac{1+x}{x} - \frac{1}{\sin x}\right) = 1.

Key concepts

Limits

In calculus, limits describe the value that a function approaches as the input approaches a certain point. In this exercise, we are dealing with the limit as \(x\) approaches 0 for the expression \((\frac{1+x}{x} - \frac{1}{\sin x})\).
Understanding how limits work is essential because they form the foundation of other concepts such as derivatives and integrals.

The limit helps us define the behavior of functions as they get very close to a specific value.

• In our case, we use limits to understand what happens to the function as x gets very close to 0.
• By examining the limit, we can simplify the expression step by step, even if the original form seems complicated.

Notice how by substituting the Maclaurin series, we make the problem more approachable to evaluate the limit.

Series Expansion

A series expansion, like the Maclaurin series, is a way to represent functions as a sum of their polynomial terms. The Maclaurin series is a special case of the Taylor series, expanded around 0.

For a function \(f(x)\), the Maclaurin series is given by:
\ \ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + ... \ \

• In our exercise, we use the Maclaurin series to expand both \(1+x\) and \(\sin x\).
• The series for \(1+x\) is quite straightforward: \(1 + x\).
• The series for \(\sin x\) includes more terms: \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\).

Using these series, we substitute them into the limit expression \((\frac{1+x}{x} - \frac{1}{\sin x})\) and simplify the terms. This simplifies the problem and makes it possible to evaluate the limit effectively.

Calculus

Calculus is a branch of mathematics that studies continuous change. It has two main branches: differential calculus and integral calculus. Limits, derivatives, and integrals are the core components of calculus.

• The concept of limits is fundamental in finding the derivative, which measures how a function changes as its input changes.
• In this exercise, we utilize limits to evaluate the behavior of the function near \(x = 0\).
• Through simplification using Maclaurin series, a fundamental technique in calculus, we successfully determine the limit.

By understanding how to use series expansion and limits, you gain deeper insights into more complex calculus problems. These tools make solving higher-level mathematical problems manageable and intuitive.