Problem 15 - Single Variable Calculus

Question

Find the indicated limit. \(\lim _{u \rightarrow-2} \sqrt[3]{\frac{3 u^{2}+2 u}{3 u^{3}-3}}\)

Step-by-Step Solution

Answer

The short answer is \(\lim _{u \rightarrow -2} \sqrt[3]{\frac{u(3u + 2)}{3(u - 1)(u^2 + u + 1)}} = \sqrt[3]{-\frac{4}{27}}\).

Step 1: Simplify the expression

First, we want to simplify the expression given inside the cube root. Simplify the numerator: \(3u^2 + 2u = u(3u + 2)\) Simplify the denominator: \(3u^3 - 3 = 3(u^3 - 1) = 3(u - 1)(u^2 + u + 1)\) Now, we rewrite the given function using these simplified expressions: \(f(u) = \sqrt[3]{\frac{u(3u + 2)}{3(u - 1)(u^2 + u + 1)}}\)

Step 2: Apply the limit

Now, let's find the limit as \(u\) approaches -2. \(\lim _{u \rightarrow-2} \sqrt[3]{\frac{u(3u + 2)}{3(u - 1)(u^2 + u + 1)}}\) Before proceeding to the next step, let's plug -2 into the simplified expression and see if it yields an indeterminate form. \(\sqrt[3]{\frac{(-2)(3(-2) + 2)}{3((-2) - 1)((-2)^2 + (-2) + 1)}} = \sqrt[3]{\frac{-4}{3(-3)(3)}} = \sqrt[3]{-\frac{4}{27}}\) As we can see, the expression is not indeterminate, so we can proceed to find the limit.

Step 3: Compute the resulting value

Now that we've found the simplified expression, we can compute the limit by simply plugging in -2 into the simplified expression. \(\lim _{u \rightarrow-2} \sqrt[3]{\frac{u(3u + 2)}{3(u - 1)(u^2 + u + 1)}} = \sqrt[3]{-\frac{4}{27}}\) The limit is equal to the cube root of the fraction \(-\frac{4}{27}\).

Key Concepts

Simplifying Expressions

Simplifying expressions is a key skill in calculus. It helps in making problems manageable and less complicated to solve. In this exercise, the expression \(\frac{3u^2 + 2u}{3u^3 - 3}\) was simplified to make it easier to compute the limit.

• The numerator, \(3u^2 + 2u\), was factored to \(u(3u + 2)\) by taking out the common factor, \(u\).
• For the denominator, \(3u^3 - 3\) was rewritten as \(3(u^3 - 1)\). It was further factored to \(3(u - 1)(u^2 + u + 1)\), utilizing the identity for the difference of cubes: \(a^3 - 1 = (a - 1)(a^2 + a + 1)\).

Simplifying expressions can also unveil possible factorizations that can cancel out terms, potentially resolving indeterminate forms. It's an essential step before diving into limit computation.

Cube Root

The cube root is crucial for transforming our understood limit into its final form. The cube root, \(\sqrt[3]{x}\), asks for the number that, when used three times in multiplication, gives you \(x\). In our exercise, after simplifying the inner function of the cube root, we were left with:

• \(\sqrt[3]{\frac{-4}{27}}\)

To evaluate this cube root, first factor \(-4\) into \(-2 \times 2\) and \(27\) into \(3 \times 3 \times 3\). The result is merely the cube root of the fraction:* \(\sqrt[3]{\left(-\frac{4}{27}\right)} = -\sqrt[3]{\left(\frac{4}{27}\right)}\) This transformation maintains the negative sign as the cube of a negative number remains negative.

Indeterminate Forms

Indeterminate forms occur when substituting a number into a function results in an undefined math expression, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It signals the need for deeper investigation to resolve the expression.In this particular problem, initially, inserting the limit value \(-2\) into our function does not produce such an indeterminate form. Instead, it simplifies the function by revealing it as a determinate value:

• \(\sqrt[3]{\left(\frac{-4}{27}\right)}\)

Finding the limit directly in this scenario becomes straightforward since the expression resolves into a clear numerical solution rather than requiring additional simplification or approaches.

Limit Computation

Limit computation focuses on analyzing how a function behaves as it approaches a particular point. In our problem, we computed the limit of a cube root function as \(u\) approaches \(-2\).During computation, we substituted the target number into the rewritten expression and calculated:

• \(\sqrt[3]{-\frac{4}{27}}\)

This process illustrates one of the central techniques in calculus: evaluating limits directly from simplified expressions when feasible. By substituting \(-2\) into the denominator and the numerator correctly formatted, we avoided complex procedures like analytical extensions or further transform.The ability to manage equations algebraically aids in quickly determining the intended limit value by directly calculating the result from the numerically simplified form.