Problem 12 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{u \rightarrow 1} \frac{(u-1)^{3}}{(1 / u)-u^{2}+3 / u-3} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Simplify the Expression in the Denominator

The denominator can be simplified as follows: \( \frac{1}{u} - u^2 + \frac{3}{u} - 3 \) can be combined by factoring out common terms. Combine the terms: \( \frac{1 + 3}{u} - (u^2 + 3) \). This simplifies to \( \frac{4}{u} - (u^2 + 3) \), or \( \frac{4 - u(u^2 + 3)}{u} \). Finally, it simplifies to \( \frac{4 - u^3 - 3u}{u} \).

Step 2: Expand and Reorganize Terms

Rewrite the simplified expression in a more straightforward form. The expression becomes \( \frac{4 - u^3 - 3u}{u} = \frac{-u^3 - 3u + 4}{u} \).

Step 3: Evaluate the Limit by Direct Substitution Initially

Substitute \( u = 1 \) directly into \( \frac{(u-1)^3}{-u^3-3u+4} = \frac{0}{0} \) which is indeterminate. We need to simplify further.

Step 4: Factorize Both Numerator and Denominator

Since both give the indeterminate form, use factoring. Numerator \( (u-1)^3 = (u-1)(u-1)(u-1) \). Denominator factorization: \( -u^3 - 3u + 4 \) can be tested for roots or factored using synthetic division or polynomial roots. The denominator factors as \( -(u-1)(u^2+u-4) \).

Step 5: Cancel Common Factors and Simplify

Cancel the common factor \( (u-1) \) from top and bottom. Simplify to get \( \frac{(u-1)^2}{-(u^2+u-4)} \).

Step 6: Direct Substitution Post-Simplification

Now that \( u = 1 \), substitute into the simplified expression \( \frac{(1-1)^2}{-(1^2+1-4)} = \frac{0}{-2} = 0 \).

Step 7: Conclude the Limit Calculation

Confirm it is no longer indeterminate after cancellation. The simplified form evaluates directly to \( 0 \) at \( u = 1 \).

Key Concepts

Limits

In calculus, a limit describes the value that a function approaches as the input approaches a given point. For the exercise, we're interested in \[ \lim_{u \rightarrow 1} \frac{(u-1)^{3}}{\frac{1}{u}-u^{2}+\frac{3}{u}-3} \]. This signifies that we want to know the behavior of the function as \( u \) gets close to 1. Limits allow us to understand the function's behavior without directly reaching the value since direct computation might not always be possible.

• Define the function you are evaluating.
• Identify if the limit can be directly substituted or if further steps are needed.

Limit calculation often involves simplifying the expression to manage indeterminate forms as seen in this task.

Indeterminate Forms

When evaluating limits, you can encounter expressions like \( \frac{0}{0} \), which indicate an indeterminate form. In this exercise, substituting \( u = 1 \) into the original expression directly results in \[ \frac{(1-1)^3}{\frac{1}{1}-1^2+\frac{3}{1}-3} = \frac{0}{0} \]. Indeterminate forms require us to manipulate the expression — through factorization or other algebraic methods — to resolve the undefined value.

• Recognize indeterminate forms by direct substitution.
• Decide on an approach to simplify the expression (e.g., factoring).

These forms are vital to understand as they highlight where more advanced techniques need to be applied.

Factorization Techniques

Factorization is a key technique for simplifying expressions, especially when faced with indeterminate forms. In this scenario, both the numerator \((u - 1)^3\) and the denominator \(-u^3 - 3u + 4\) need factoring for simplification.

• For the numerator, factoring follows as \((u-1)(u-1)(u-1)\).
• For the denominator, we apply polynomial factorization, recognizing that it factors as \(-(u-1)(u^2+u-4)\).

Factorization allows us to cancel common factors, simplifying the limit expression.

Direct Substitution

Direct substitution is often the simplest method for evaluating limits when the expression can be simplified to avoid indeterminate forms. Once the indeterminate form is eliminated by canceling out the common factor \((u-1)\), you can directly substitute \(u = 1\) into the simplified expression \( \frac{(u-1)^2}{-(u^2+u-4)}. \) After substitution, the solution evaluates to \( \frac{0}{-2} = 0. \)Direct substitution, when viable, provides a straightforward path to calculating limits.

• Check if simplification allows direct substitution.
• Use substitution to reach the limit value without further complications.

It's an essential part of limit evaluation, where simplification methods make it applicable.