Question

Use a calculator to graph the function and estimate the value of the limit, then use L'Hôpital's rule to find the limit directly. $$ \text { [T] } \lim _{x \rightarrow 0}\left(\csc x-\frac{1}{x}\right) $$

Short answer

The limit is 0.

Step-by-step solution

Graph the Function

To graph \( f(x) = \csc(x) - \frac{1}{x} \), use a calculator or graphing software. You will see that as \( x \rightarrow 0 \), the graph seems to approach a particular value. However, keep in mind that both \( \csc(x) \) and \( \frac{1}{x} \) have discontinuities at \( x = 0 \). From the graph, the limit appears to be around 0.

Confirm Indeterminate Form

Evaluate the limit expression to confirm it's an indeterminate form. As \( x \rightarrow 0 \), \( \csc(x) = \frac{1}{\sin(x)} \) and \( \frac{1}{x} \) both tend to infinity, so we have \( \lim_{x \rightarrow 0} \left( \frac{1}{\sin(x)} - \frac{1}{x} \right) \), which results in the form \( \frac{\infty - \infty}{1} \). This is not directly solvable, suggesting the use of L'Hôpital's rule.

Rewrite for L'Hôpital's Rule

Rewrite the expression so it has a common denominator: \( \csc(x) - \frac{1}{x} = \frac{1}{\sin(x)} - \frac{1}{x} = \frac{x - \sin(x)}{x \sin(x)} \). Verify that this is indeed an indeterminate form \( \frac{0}{0} \) as \( x \rightarrow 0 \).

Apply L'Hôpital's Rule

Apply L'Hôpital's rule by differentiating the numerator and the denominator. The derivative of the numerator \( x - \sin(x) \) is \( 1 - \cos(x) \), and the derivative of the denominator \( x \sin(x) \) is \( \sin(x) + x \cos(x) \). Thus, \[ \lim_{x \rightarrow 0} \frac{x - \sin(x)}{x \sin(x)} = \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{\sin(x) + x \cos(x)} \].

Evaluate the Limit

Evaluate the limit \( \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{\sin(x) + x \cos(x)} \). As \( x \) approaches 0, the numerator \( 1 - \cos(x) \) approaches 0, and the denominator \( \sin(x) + x \cos(x) \) approaches 0. Apply L'Hôpital's Rule again if necessary or use Taylor expansions for sinus and cosine to evaluate precisely. This results in evaluating the Taylor expansions: leading to the evaluation reaching 0.

Key concepts

Indeterminate Forms

In calculus, an indeterminate form arises when we try to evaluate limits that initially seem undefined or ambiguous. A common example of an indeterminate form is when we have expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In our exercise, we identified an indeterminate form when calculating the limit \( \lim_{x \rightarrow 0} \left( \csc x - \frac{1}{x} \right) \). As \( x \) approaches zero, both \( \csc(x) = \frac{1}{\sin(x)} \) and \( \frac{1}{x} \) tend towards infinity.

Initially, this produces an indeterminate form \( \frac{\infty - \infty}{1} \). By rewriting the expression to \( \frac{x - \sin(x)}{x \sin(x)} \), we identified a \( \frac{0}{0} \) form, allowing us to apply L'Hôpital's rule to resolve it. Indeterminate forms often signal the need for more advanced techniques to be used for accurate limit evaluation.

Graphing Functions

Graphing functions is an essential skill in calculus, helping us visualize and estimate the behavior of functions as they approach certain points. For the function \( f(x) = \csc(x) - \frac{1}{x} \), graphing helps us initially assess the limit as \( x \rightarrow 0 \).

Using a graphing calculator, we can visually confirm whether the function approaches a specific value near \( x = 0 \). By examining the graph, students can estimate that the limit seems to stabilize around a particular number. However, because both \( \csc(x) \) and \( \frac{1}{x} \) have discontinuities at \( x = 0 \), the graph might appear erratic very close to the point.

Graphing is a valuable first step for understanding the behavior of complex functions and confirming the presence of an indeterminate form before using analytical methods such as L'Hôpital's Rule.

Limit Evaluation

Evaluating limits is a fundamental task in calculus, crucial for understanding function behavior near points of interest. The limit \( \lim_{x \rightarrow 0} \left( \csc(x) - \frac{1}{x} \right) \) illustrates the need to carefully evaluate expressions that initially appear undefined.

We first confirm the presence of an indeterminate form, such as \( \frac{\infty - \infty}{1} \) at \( x = 0 \). By rewriting this in the form \( \frac{0}{0} \), we prepare it for L'Hôpital's Rule:

• Differentiate the numerator and the denominator separately.
• Use the derivatives to find the limit as \( x \rightarrow 0 \).

This technique mirrors how we apply derivative understanding to solve limits. If the first application of L'Hôpital's Rule still results in an indeterminate form, try applying the rule again or use other techniques, such as Taylor expansions, for a precise evaluation.

Taylor Expansions

Taylor expansions provide a powerful method for approximating functions using polynomial expressions, especially when exact evaluation is challenging. When evaluating the limit \( \lim_{x \rightarrow 0} \left( \csc(x) - \frac{1}{x} \right) \), Taylor series can simplify solving indeterminate forms.

For functions like \( \sin(x) \) and \( \cos(x) \), we use their Taylor series around zero.

• \( \sin(x) \approx x - \frac{x^3}{6} \)
• \( \cos(x) \approx 1 - \frac{x^2}{2} \)

Substituting these into our expression \( \frac{1 - \cos(x)}{\sin(x) + x\cos(x)} \), simplifies to more manageable terms, allowing us to accurately compute the limit.

These expansions illustrate how complex trigonometric functions behave near specific points, like zero, honing in on key features with polynomials. Taylor expansions not only aid in limits but broadly in approximating difficult functions across calculus.