Question

Use analytical methods to evaluate the following limits. $$\lim _{x \rightarrow 6} \frac{\sqrt[5]{5 x+2}-2}{1 / x-1 / 6}$$

Short answer

Question: Determine the limit of the given expression as x approaches 6. Expression: $$\lim_{x \rightarrow 6} \frac{\sqrt[5]{5x+2}-2}{\frac{1}{x}-\frac{1}{6}}$$ Answer: $$\lim_{x \rightarrow 6}\frac{5}{\frac{-1}{36}(122)}=-\frac{180}{122}$$

Step-by-step solution

Identify the problem

Our goal is to find the limit of the given expression as \(x\) approaches 6. $$\lim_{x \rightarrow 6} \frac{\sqrt[5]{5x+2}-2}{\frac{1}{x}-\frac{1}{6}}$$

Rationalize the denominator

First, we are going to rationalize the denominator to make it easier to work with. To do this, we find a common denominator for the two fractions in the denominator and combine them: $$\lim_{x \rightarrow 6} \frac{\sqrt[5]{5x+2}-2}{\frac{6-x}{6x}}$$

Rationalize the numerator

Now, we need to find a way to deal with the fifth root expression. It would be helpful to rationalize the numerator as well. We can do that by multiplying and dividing the expression by its conjugate. $$\lim_{x \rightarrow 6} \frac{(\sqrt[5]{5x+2}-2)(\sqrt[5]{(5x+2)^4(5x+2)}+\sqrt[5]{(5x+2)^3(5x+2)^2}+...+\sqrt[5]{(5x+2)^2(5x+2)}+2)}{(\frac{6-x}{6x})(\sqrt[5]{(5x+2)^4(5x+2)}+\sqrt[5]{(5x+2)^3(5x+2)^2}+...+\sqrt[5]{(5x+2)^2(5x+2)}+2)}$$

Simplify the numerator

When we multiply the two expressions in the numerator, it follows the pattern of the difference of cubes, which can be simplified: $$\lim_{x \rightarrow 6} \frac{(5x+2)-2^5}{(\frac{6-x}{6x})(\sqrt[5]{(5x+2)^4(5x+2)}+\sqrt[5]{(5x+2)^3(5x+2)^2}+...+\sqrt[5]{(5x+2)^2(5x+2)}+2)}$$ Simplify the numerator further: $$\lim_{x \rightarrow 6} \frac{5x-30}{(\frac{6-x}{6x})(\sqrt[5]{(5x+2)^4(5x+2)}+\sqrt[5]{(5x+2)^3(5x+2)^2}+...+\sqrt[5]{(5x+2)^2(5x+2)}+2)}$$

Cancel out terms

As we have simplified both the numerator and denominator, we can now cancel out some terms: $$\lim_{x \rightarrow 6} \frac{5(x-6)}{(\frac{-1(x-6)}{6x})(\sqrt[5]{(5x+2)^4(5x+2)}+\sqrt[5]{(5x+2)^3(5x+2)^2}+...+\sqrt[5]{(5x+2)^2(5x+2)}+2)}$$ Now, we can cancel out the \((x-6)\) term from both the numerator and the denominator: $$\lim_{x \rightarrow 6} \frac{5}{(\frac{-1}{6x})(\sqrt[5]{(5x+2)^4(5x+2)}+\sqrt[5]{(5x+2)^3(5x+2)^2}+...+\sqrt[5]{(5x+2)^2(5x+2)}+2)}$$

Evaluate the limit

Now, as the expression is considerably simplified, we can substitute \(x=6\) in the equation: $$\lim_{x \rightarrow 6} \frac{5}{(\frac{-1}{6(6)})(\sqrt[5]{(5(6)+2)^4(5(6)+2)}+\sqrt[5]{(5(6)+2)^3(5(6)+2)^2}+...+\sqrt[5]{(5(6)+2)^2(5(6)+2)}+2)}$$ After substituting the value of \(x\) and simplifying, we obtain the final answer: $$\lim_{x \rightarrow 6}\frac{5}{\frac{-1}{36}(122)}=\boxed{-\frac{180}{122}}$$

Key concepts

Rationalizing the Numerator

Rationalizing is a useful algebraic technique aimed at eliminating roots or radicals from fractions, usually to simplify the process of finding limits. In this context, it involves removing the root from the numerator. Here's how you typically approach it:

• Identify the radical in the numerator, in our case, \( \sqrt[5]{5x+2} - 2 \).
• Multiply both the numerator and the denominator by the conjugate of the numerator. The conjugate of \( a - b \) is \( a + b \).

This approach exploits the identity \((a-b)(a+b) = a^2-b^2 \), helping us clear out the root. It’s crucial because it allows further simplification by creating terms that are much easier to handle when calculating limits.

Difference of Fifth Powers

The term "difference of fifth powers" isn't as commonly seen as its simpler counterparts like "difference of squares." Nevertheless, it follows a similar principle but for higher powers.

• It involves expressions of the form \((a-b)\), when raised to the fifth power.
• The expansion follows patterns that can be simplified to make further calculations easier.

For the exercise, treating \( \sqrt[5]{5x+2} - 2 \) as a simplified difference of fifth powers helps us recognize pattern-based similarities and approach further simplification using algebraic identities that reveal underlying factors. This simplification is vital to tackling polynomial roots effectively.

Simplifying Expressions

Simplifying is key to evaluating limits as it reduces complex expressions to simpler forms that are easier to manage. In this case:

• Begin by simplifying the fraction in the denominator, turning it into a more manageable operation using common denominators.
• Apply techniques like canceling common terms between the numerator and denominator.
• Simplify any remaining expressions, looking for opportunities to substitute the limit point \(x = 6\).

This process not only assists with clarity but also ensures every step logically leads to a point where direct substitution in the limit becomes viable, allowing us to confidently evaluate what becomes of the expression as \( x \) approaches the specific value.