Question

Evaluate the following limits. $$\lim _{x \rightarrow 2} \frac{\sqrt[3]{3 x+2}-2}{x-2}$$

Short answer

$$\lim _{x \rightarrow 2} \frac{\sqrt[3]{3 x+2}-2}{x-2}$$ Answer: The limit of the expression as x approaches 2 is: $$\lim _{x \rightarrow 2} \frac{\sqrt[3]{3 x+2}-2}{x-2} = 12 + 2\sqrt[3]{32}$$

Step-by-step solution

Direct Substitution

To start, plug in the value x=2 and check if the limit is in the indeterminate form or not. $$\frac{\sqrt[3]{3(2)+2}-2}{2-2} = \frac{\sqrt[3]{8}-2}{0} = \frac{2 - 2}{0} = \frac{0}{0}$$ Since the expression is in the indeterminate form, we need to manipulate the expression. Step 2: Factor the numerator

Factor the numerator

Let's find another expression for the numerator, so that we can cancel out the denominator. We will do that using the difference of cubes formula: $$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$ So, our expression becomes: $$\lim _{x \rightarrow 2} \frac{(\sqrt[3]{3 x+2}-2)\big((\sqrt[3]{3 x+2})^2 + (\sqrt[3]{3 x+2})(2) + 2^2\big)}{x-2}$$ Step 3: Simplify the numerator

Simplify the numerator

Now simplify the numerator expression: $$\big((\sqrt[3]{3 x+2})^2 + (\sqrt[3]{3 x+2})(2) + 2^2\big) = (3x + 2 + 2\sqrt[3]{(3x+2)4} + 4)$$ Now, it simplifies to: $$(3x + 6 + 2\sqrt[3]{(3x+2)4})$$ Step 4: Cancel terms and evaluate the limit

Cancel terms and evaluate the limit

Now, we cancel out the \((x-2)\) term in the numerator and denominator: $$\lim _{x \rightarrow 2} \frac{3x + 6 + 2\sqrt[3]{(3x+2)4}}{1}$$ Evaluate the limit by direct substitution: $$\lim _{x \rightarrow 2} (3(2) + 6 + 2\sqrt[3]{(3(2)+2)4}) = 6 + 6 + 2\sqrt[3]{(6+2)4} = 12 + 2\sqrt[3]{32}$$ So, the limit is: $$\lim _{x \rightarrow 2} \frac{\sqrt[3]{3 x+2}-2}{x-2} = 12 + 2\sqrt[3]{32}$$

Key concepts

Indeterminate Forms

When dealing with limits in calculus, you might come across expressions that are undefined or confusing at first glance. These are known as indeterminate forms. A common example is when substituting a number into a limit expression gives a form like \( \frac{0}{0} \). This occurs because both the numerator and the denominator evaluate to zero, leading to an undefined situation.
An indeterminate form does not mean that the limit doesn't exist, but rather that you have to do some extra work to find it. To resolve this, you need to manipulate the expression to eliminate the indeterminate form. This could involve algebraic tricks, such as factoring, or special techniques like L'Hopital's Rule, depending on the complexity of the function.
In our example, direct substitution led to \( \frac{0}{0} \), prompting us to rewrite the expression more aptly.

Difference of Cubes

The difference of cubes is a useful algebraic formula that helps simplify expressions involving cubed terms. The formula is: \[a^3 - b^3 = (a-b)(a^2 + ab + b^2)\]
This formula is handy for breaking down complex expressions, especially those linked to cubic roots or exponents. By recognizing terms that resemble \(a^3 - b^3\), you can transform a complex expression into simpler factors, often allowing you to cancel terms and simplify further.
In the solved exercise, the expression \( \sqrt[3]{3x+2} - 2 \) can be thought of as \( a - b \) where \( a = \sqrt[3]{3x+2} \) and \( b = 2 \). By applying the difference of cubes formula, we simplify the numerator to make the problematic denominator easier to handle.

Direct Substitution

Direct substitution is often the first step when evaluating limits. It involves simply plugging the value, to which \(x\) approaches, directly into the function. This method is straightforward and ideal for cases where substitution yields a proper numerical result directly.
However, when direct substitution results in an indeterminate form, as it did in our example with \( \frac{0}{0} \), it becomes clear that further simplification or manipulation is necessary. Only by overcoming these special cases can the true limit be evaluated accurately.
Direct substitution showcases its power in simple scenarios, but it also serves as a crucial indicator for complex instances, guiding the need to employ additional algebraic techniques.

Simplifying Expressions

Simplifying expressions is a vital skill in mathematics, particularly in limit problems. The goal is to rewrite an expression in a more straightforward form, often to allow substitution or cancellation of terms. This process invariably involves algebraic manipulation, such as factorization, expansion, or conjugation.
In the given exercise, simplifying the initial expression was essential. By applying the difference of cubes formula, the expression was reorganized in such a way that it removed the indeterminate form. Once simplified, canceling equivalent terms from the numerator and denominator becomes possible.
The revised function is easily substitute-able, as seen when we simplified and then evaluated the limit by substituting \(x = 2\). This simplification ensures that once-direct substitution now yields an actual numerical limit, resolving the problem efficiently.