Question

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

Short answer

$\underset{x\to 0}{lim}({\text{csc}}^{2}x-\frac{1}{x^2})=\frac{1}{3}$

Step-by-step solution

Recall Maclaurin series for relevant functions

The Maclaurin series of $\text{csc}x$ can be deduced from the series of $\text{sin}x$. Recall that $\text{sin}x=x-\frac{x^3}{6}+O(x^5)$. Hence, $\text{csc}x=\frac{1}{\text{sin}x}\approx \frac{1}{x-\frac{x^3}{6}+O(x^5)}$.

Simplify the series for $\text{csc}x$

To approximate $\text{csc}x$, invert the Maclaurin series of $\text{sin}x$. Therefore, $\text{csc}x=\frac{1}{x}+\frac{x}{6}+O(x^3)$.

Find the series for ${\text{csc}}^{2}x$

Square the result from Step 2: $${\text{csc}}^{2}x={(\frac{1}{x}+\frac{x}{6}+O(x^3))}^2=\frac{1}{x^2}+\frac{2}{6}+O(x^2).$$

Express the limit using the series

The given limit can now be written as: $${\text{csc}}^{2}x-\frac{1}{x^2}=(\frac{1}{x^2}+\frac{1}{3}+O(x^2))-\frac{1}{x^2}=\frac{1}{3}+O(x^2).$$

Evaluate the limit as \ x \ approaches 0

As \ x \ approaches 0, the higher order terms $O(x^2)$ vanish, leaving $$\frac{1}{3}.$$ Therefore, \ \lim_{x \rightarrow 0} \left( \text{csc}^2 \ x - \frac{1}{x^2} \right) = \frac{1}{3}.

Key concepts

Maclaurin Series

The Maclaurin series is a special case of the Taylor series, expanded around zero. It's a polynomial approximation of functions centered at zero. For instance, the Maclaurin series for $\text{sin}x$ is: $$\text{sin}x=x-\frac{x^3}{6}+O(x^5)$$.
What this means is that we approximate $\text{sin}x$ using its powers series expansion up to higher-order terms (termed as $O(x^5)$). This makes it easier to evaluate functions, especially when dealing with limits. When applying Maclaurin series, breaking down functions to their power series form simplifies complex expressions, letting us isolate and evaluate key terms, especially around zero.
Understanding these series helps tremendously with functions that are difficult to handle directly.

Cosecant (csc) Function

The cosecant function, denoted as $\text{csc}x$, is the reciprocal of the sine function: \text{csc} \ x \ = \frac{1}{\text{sin} \ x}\.
Its behavior near zero is particularly interesting because $\text{sin}x$ approaches zero, making $\text{csc}x$ grow very large.
To understand $\text{csc}x$ near zero using series expansions, we convert $\text{sin}x$ into its Maclaurin series form and invert it. For example, from the given exercise, we have: $\text{sin}x=x-\frac{x^3}{6}+O(x^5)$ leading to: $\text{csc}x=\frac{1}{x}+\frac{x}{6}+O(x^3)$.
This simplification allows us to more easily manage complex operations, such as squaring or differentiating.

Limit Evaluation

Limits give us the value a function approaches as the input approaches some value. Evaluating limits involving complicated functions is more straightforward using series expansions.
In this exercise, we want to find: ${\text{lim}}_{x\to 0}({\text{csc}}^{2}x-\frac{1}{x^2})$. We start by using the Maclaurin series: $\text{csc}x=\frac{1}{x}+\frac{x}{6}+O(x^3)$. Squaring it, we get: ${\text{csc}}^{2}x=(\frac{1}{x}+\frac{x}{6}+O(x^3){)}^2=\frac{1}{x^2}+\frac{1}{3}+O(x^2)$.
This form makes it easy to evaluate the limit as the higher-order terms disappear. It shows that ${\text{csc}}^{2}x-\frac{1}{x^2}\to \frac{1}{3}$.
Breaking the functions down with series reveals the value the limit approaches.

Taylor Series

A Taylor series is a representation of a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. When this point is zero, it's termed a Maclaurin series. The general form of a Taylor series for a function $f(x)$ is:
$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\dots$$
In our example, we've used the Maclaurin series, where $a=0$. The Taylor series for $\text{sin}x$ around zero is:
$\text{sin}x=x-\frac{x^3}{6}+O(x^5)$
This series helps simplify the evaluation of limits involving trigonometric functions. The series expansion approximates these functions, facilitating their manipulation and limit evaluation by providing a clear polynomial form that captures their behavior near the point of expansion.