Question

Use Maclaurin series to evaluate: $\underset{x\to 0}{lim}(\frac{1+x}{x}-\frac{1}{\sin x})$

Short answer

The limit is 1.

Step-by-step solution

Recall the Maclaurin series for basic functions

Use the Maclaurin series expansions for $\frac{1+x}{x}$ and $\frac{1}{\sin x}$. The Maclaurin series for $\frac{1+x}{x}$ is $\frac{1}{x}+1$ since $\frac{1+x}{x}=\frac{1}{x}+1$. The Maclaurin series for $\frac{1}{\sin x}$ requires the series for $\sin x$ which is $x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^7)$. Hence, $\frac{1}{\sin x}$ can be approximated by $\frac{1}{x}+\frac{x}{6}+O(x^3)$.

Express the difference using the Maclaurin series

Rewrite $\frac{1+x}{x}-\frac{1}{\sin x}$ as $\frac{1}{x}+1-\frac{1}{x}-\frac{x}{6}$. By substituting the series we derived, we have $\frac{1+x}{x}$ as $\frac{1}{x}+1$ and $\frac{1}{\sin x}$ as $\frac{1}{x}+\frac{x}{6}$. Thus the expression is $\frac{1}{x}+1-\frac{1}{x}-\frac{x}{6}$.

Simplify the expression

Notice that $\frac{1}{x}$ terms cancel each other. We are left with $1-\frac{x}{6}$.

Evaluate the limit as x approaches 0

To find the limit as $x\to 0$, we substitute $x=0$ into the simplified expression from Step 3. Hence, the limit is $1-\frac{0}{6}=1$.

Key concepts

Limit Evaluation

In calculus, limit evaluation is a powerful tool. It helps us find the value that a function approaches as the input approaches a particular point. For the problem given, we need to evaluate $\underset{x\to 0}{lim}(\frac{1+x}{x}-\frac{1}{\sin x})$. Limits can often be tricky, but using series expansions like Maclaurin series can simplify them greatly. By substituting a complex function with its series expansion, we can sometimes make limits far easier to evaluate. Let's dive deeper into series expansions to see why they are particularly useful here.

Series Expansions

Series expansions allow us to represent functions as infinite sums of terms. The Maclaurin series is a type of series expansion that approximates a function around 0. For instance, the Maclaurin series for $x$ is $x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^7)$. This series can provide a better understanding of how $\sin x$ behaves near zero. Transforming functions into series form helps to break them down into simpler components. In our exercise, we utilized these expansions for both $\frac{1+x}{x}$ and $\frac{1}{\sin x}$. By rewriting these functions in their series forms, we simplified our original limit problem, leading to a much easier calculation.

Trigonometric Series

Trigonometric functions like $\sin x$ have specific series expansions that can be super handy for limit calculations. The Maclaurin series for sine is especially useful because it gives us a polynomial form for $\sin x$:

• $\sin x\approx x-\frac{x^3}{6}+O(x^5)$

Expanding a trigonometric function can reduce complicated expressions into simpler terms. For example, in the original exercise, knowing the Maclaurin series of $\sin x$ allowed us to approximate $\frac{1}{\sin x}$. By breaking down these trigonometric functions, we made it possible to subtract and simplify effectively.

Calculus

Calculus forms the foundation for exploring limits, derivatives, and integrals. The given problem requires a limit evaluation, closely tied to the concept of derivatives. Understanding how to manipulate functions using calculus, especially through series expansions, is crucial. In a step-by-step manner we:
* Substituted functions with their Maclaurin series.
* Simplified the series to get rid of complex terms.
* Evaluated the edge-case as $x$ approaches 0.
Mastering these fundamental techniques can vastly improve your problem-solving skills in calculus. Hence, the limit was evaluated as 1 by substituting the simplified result back into the original expression.