Question

Use Maclaurin series to evaluate the limits. $\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{\cos x}{{\sin }^{2}x})$

Short answer

The limit is 0.

Step-by-step solution

Recall the Maclaurin series for sine and cosine

The Maclaurin series for $\sin x$ and $\cos x$ are essential for solving this problem. Recall that: $\sin x=x-\frac{x^3}{6}+O(x^5)$ and $\cos x=1-\frac{x^2}{2}+O(x^4)$.

Expand ${\sin }^{2}x$ using Maclaurin series

Using the Maclaurin series for $\sin x$: ${\sin }^{2}x={(x-\frac{x^3}{6}+O(x^5))}^2=x^2-\frac{x^4}{3}+O(x^6)$.

Expand $\cos x$ using Maclaurin series

Using the Maclaurin series for $\cos x$:$\cos x=1-\frac{x^2}{2}+O(x^4)$.

Substitute the expansions into the limit

Substitute the Maclaurin series expansions into the original limit expression: $\frac{1}{x^2}-\frac{\cos x}{{\sin }^{2}x}=\frac{1}{x^2}-\frac{1-\frac{x^2}{2}+O(x^4)}{x^2-\frac{x^4}{3}+O(x^6)}$.

Simplify the limit expression

Simplify the expression step by step: $=\frac{1}{x^2}-\frac{1-\frac{x^2}{2}+O(x^4)}{x^2(1-\frac{x^2}{3}+O(x^4))}=\frac{1}{x^2}-\frac{1-\frac{x^2}{2}+O(x^4)}{x^2-\frac{x^4}{3}+O(x^6)}$.\ When simplifying further, notice the higher-order terms approach zero as x approaches zero.

Evaluate the limit

$\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{1-\frac{x^2}{2}+O(x^4)}{x^2-\frac{x^4}{3}+O(x^6)})=\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{1}{x^2})=0$.

Key concepts

Taylor series expansion

The Taylor series expansion is a powerful mathematical tool that allows us to express functions as infinite sums of their derivatives around a specific point, usually zero. This expansion helps in approximating functions for small input values. The Maclaurin series is a special case of the Taylor series, centered at zero. For example, the Maclaurin series for $\sin x$ and $\cos x$ are crucial for solving calculus problems involving trigonometric functions.$\begin{array}{rlr}\sin x& =x-\frac{x^3}{6}+O(x^5),\cos x& =1-\frac{x^2}{2}+O(x^4)\end{array}$
We use this expansion to approximate $\sin x$ and $\cos x$ near zero.\

Limit evaluation

Evaluating limits, especially as a variable approaches zero, is a common problem in calculus. When direct substitution results in an indeterminate form, we can use series expansions to simplify and evaluate the limit. In the given problem, we use the Maclaurin series for trigonometric functions to break down the complex expressions into simpler forms.
This approach transforms the original expression into something manageable, allowing us to evaluate the limit accurately.
Here, we expand ${\sin }^{2}x$ and $\cos x$ using their series expansions and substitute these into the given limit expression. Simplifying step by step, and recognizing that higher-order terms become negligible as $x$ approaches zero, enables us to find the limit precisely.

Trigonometric functions

Trigonometric functions like $\sin x$ and $\cos x$ are fundamental in calculus, and their properties and expansions are vital to many problems. Understanding their series expansions helps in solving limits, derivatives, and integrals.
In our problem, the Maclaurin series for these functions provide a simpler way to handle the limit. $\sin x$ is approximated as $x-\frac{x^3}{6}+O(x^5)$, and $\cos x$ as $1-\frac{x^2}{2}+O(x^4)$. Using these expansions, we can simplify complex expressions and evaluate them near zero.
This method is particularly useful when dealing with indeterminate forms that are common in limits involving trigonometric functions.

Calculus techniques

Effective use of calculus techniques is crucial for solving advanced math problems. Techniques like series expansions, differentiation, and limit evaluation are interconnected tools that enable us to handle complex expressions efficiently. In this example, the Maclaurin series for trigonometric functions allows us to break down and simplify expressions, making the evaluation of limits more straightforward.
Moreover, the process of simplifying and substituting higher-order terms showcases the elegance and power of calculus in solving real-world problems. Mastering these techniques enhances problem-solving skills and deepens understanding of mathematical concepts.