Question

Use a CAS to evaluate the limits. $$ \lim _{x \rightarrow 0} \frac{1-\cos \left(x^{2}\right)}{x^{3} \sin x} $$

Short answer

The limit is \( \frac{1}{2} \).

Step-by-step solution

Substitute x with 0 Directly

First, we attempt a direct substitution of the limit into the expression. Substituting 0 for \(x\), we have: \( \lim _{x \rightarrow 0} \frac{1- \cos(x^2)}{x^3 \sin x} = \frac{1 - \cos(0)}{0^3 \sin(0)} = \frac{0}{0} \), which is an indeterminate form.

Use Series Expansions for Trigonometric Functions

To resolve the indeterminate form, use the Taylor series expansions for \( \cos(x^2) \) and \( \sin x \). We know that \( \cos(x^2) \approx 1 - \frac{(x^2)^2}{2} = 1 - \frac{x^4}{2} \) and \( \sin x \approx x - \frac{x^3}{6} \). These approximations hold true as \(x\) approaches 0.

Substitute Series Expansions

Substitute the series expansions into the limit expression: \[ \lim _{x \rightarrow 0} \frac{1- (1 - \frac{x^4}{2})}{x^3 (x - \frac{x^3}{6})} = \lim _{x \rightarrow 0} \frac{\frac{x^4}{2}}{x^4 - \frac{x^6}{6}} \].

Simplify the Expression

Simplifying \( \frac{x^4}{2x^4 - \frac{x^6}{6}}\), we factor out \(x^4\) from the denominator: \[ \lim _{x \rightarrow 0} \frac{\frac{x^4}{2}}{x^4(1 - \frac{x^2}{6})} = \lim _{x \rightarrow 0} \frac{1}{2(1 - \frac{x^2}{6})} \].

Evaluate the Limit

Finally, as \(x\) approaches 0, \(1 - \frac{x^2}{6}\) approaches 1. Thus, evaluate the limit directly: \(\lim _{x \rightarrow 0} \frac{1}{2(1 - \frac{x^2}{6})} = \frac{1}{2} \).

Key concepts

Limits

Limits are a fundamental concept in calculus, used to explore the behavior of functions as they approach specific points. Imagine you're driving towards a red traffic light. The limit is essentially where you will end up as you nearly reach the stop line. In mathematics, we want to know what value a function approaches as the input moves closer and closer to a certain point. This helps determine the function's behavior near that point, even if it's not defined exactly there.

In our example, we looked at the limit of \[\lim _{x \rightarrow 0} \frac{1-\cos(x^2)}{x^3 \sin x}\]When we substituted 0, we ran into an indeterminate form \(\frac{0}{0}\), which means the limit can't be directly evaluated with simple substitution. But, yes, we can solve it by other means like using Taylor series, which we'll explain next.

Taylor series

The Taylor series is like a mathematician's secret weapon. It allows us to approximate complex functions with simpler polynomials. By expanding functions as infinite sums of terms calculated from the values of their derivatives at a single point, usually around \(x = 0\) (known as a Maclaurin series), you can transform tricky functions into something more manageable.

In our case, to tackle the indeterminate form, we took the Taylor series expansion for \(\cos(x^2)\) and \(\sin x\).
- For \(\cos(x^2)\), it became \(1 - \frac{x^4}{2}\), since higher powers become insignificant as \(x\) heads toward zero.- For \(\sin x\), it was \(x - \frac{x^3}{6}\). These series make it easier to simplify the expression and evaluate the limit without facing the problematic \(\frac{0}{0}\) situation.

Indeterminate forms

Indeterminate forms are a bit like trying to solve a riddle when there seems to be no obvious answer. These take shapes like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), where direct calculation gives a meaningless result.

In limit problems, they indicate uncertainty, exact behavior of the function is unclear, and additional techniques are needed. Often, methods like Taylor series or L'Hôpital's rule are used here to resolve these forms and find a definitive answer.

For our original problem, encountering \(\frac{0}{0}\) led us to use the Taylor series to find a limit that's meaningful and calculable.

Trigonometric functions

Trigonometric functions like \(\sin x\) and \(\cos x\) are critical in calculus. They describe the relationships of angles and sides in triangles, extending those ideas to all of mathematics with periodic behavior.

In limits and series, these functions are often involved and have well-known expansions. For small angles, or as \(x\) approaches zero, \(\sin x\) can be approximated by \(x\) and \(\cos x\) closely follows 1 minus a small squared value.

• For \(\sin x\), use the approximation \(x - \frac{x^3}{6}\).
• For \(\cos x\), use \(1 - \frac{x^4}{2}\). This helps simplify expressions like the one in our problem.

Knowing these approximations allows us to break down complex expressions into simpler problems, especially useful when using Taylor series to tackle indeterminate limits.