Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{x^{3}-3 x^{2}+x}{x^{3}-2 x} $$

Short answer

The limit is -1/2.

Step-by-step solution

Evaluate the limit directly

First, substitute the value of \(x = 0\) into the given fraction to check if it leads to an indeterminate form. Substituting \(x = 0\) into the expression \(\frac{x^3 - 3x^2 + x}{x^3 - 2x}\) gives: \(\frac{0^3 - 3(0)^2 + 0}{0^3 - 2(0)} = \frac{0}{0}\). This result is an indeterminate form, so we can consider using l'Hôpital's Rule.

Confirm condition for l'Hôpital's Rule

l'Hôpital's Rule can be used when the limit results in an indeterminate form of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Here, the form is \(\frac{0}{0}\) after direct substitution, confirming our ability to apply l'Hôpital's Rule.

Differentiate the numerator and denominator

To apply l'Hôpital's Rule, we differentiate the numerator \(x^3 - 3x^2 + x\) and denominator \(x^3 - 2x\) separately. - Differentiating the numerator: \( \frac{d}{dx}(x^3 - 3x^2 + x) = 3x^2 - 6x + 1\) - Differentiating the denominator: \( \frac{d}{dx}(x^3 - 2x) = 3x^2 - 2\)

Evaluate the new limit

Now, substitute again \(x = 0\) into the new fraction obtained from the derivatives: \[ \lim _{x \rightarrow 0} \frac{3x^2 - 6x + 1}{3x^2 - 2} \] Substituting \(x = 0\), we get: \(\frac{3(0)^2 - 6(0) + 1}{3(0)^2 - 2} = \frac{1}{-2}\). This results in \(-\frac{1}{2}\), which is not an indeterminate form.

Key concepts

Indeterminate Forms

When we explore the world of limits, we often encounter a mysterious entity known as indeterminate forms. These occur when trying to evaluate the limit leads to an expression that doesn't immediately reveal any useful information, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). For example, in our problem, when substituting \( x = 0 \) into \( \frac{x^3 - 3x^2 + x}{x^3 - 2x} \), we obtain \( \frac{0}{0} \). This is a classic indeterminate form, and it signals that the direct path to finding the limit isn't straightforward. Rather than a dead end, indeterminate forms invite us to explore deeper—often using methods like algebraic manipulation or the powerful tool of l'Hôpital's Rule.

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. It's the process of determining the rate at which a function is changing at any given point. In the context of l'Hôpital's Rule, differentiation comes into play to help resolve limits involving indeterminate forms. - For the numerator of our example, \( x^3 - 3x^2 + x \), the derivative is computed as \( 3x^2 - 6x + 1 \).- Meanwhile, for the denominator, \( x^3 - 2x \), the derivative is \( 3x^2 - 2 \).By differentiating the top and bottom of our fraction separately, we create a new ratio that can potentially simplify the evaluation of our limit. This refreshed fraction is crucial for the next steps in tackling the limit problem where indeterminate forms are involved.

Calculating Limits

Calculating limits is about understanding the behavior of a function as the input approaches a particular value. Once we recognize the presence of an indeterminate form, our task becomes determining a strategy to find the limit without using direct substitution, which often leads to this ambiguity. The approach typically involves:

• Recognizing the limit type (like \( \frac{0}{0} \)).
• Differentiate the involved functions if l'Hôpital's Rule is applicable.
• Substitute back or simplify to arrive at a value.

In practice, successfully calculating the limit involves these thoughtful considerations and applying rules deftly to achieve a non-indeterminate result. Ultimately, like we did in this exercise, it ensures that the solution emerges logically, bypassing the numerical stalemate the indeterminate forms present.

Direct Substitution

Direct substitution is often the first step when evaluating limits. It involves substituting the value that \( x \) approaches directly into the expression. If the substitution yields a sensible number, the limit is resolved right away. However, when direct substitution leads to something ambiguous like \( \frac{0}{0} \), as with our problem, it's clear that another strategy must be employed.Direct substitution is crucial initially because:

• It affirms if a limit is straightforwardly evaluable or if deeper analysis is necessary.
• Helps in identifying critical points where the function might have discontinuities or unusual behavior.

When it fails, like in this exercise, it's an essential diagnostic tool that signals the need for expanded mathematical techniques, such as l'Hôpital's Rule, to progress.