We are tasked with evaluating the following limit:

$\lim_{x \to 3} \frac{\sqrt{3x + 40} - 7}{(5x - 7)^{1/3} - 2}.$

Step 1: Direct Substitution

Let's substitute $x = 3$ into the numerator and denominator:

• Numerator: $\sqrt{3(3) + 40} - 7 = \sqrt{49} - 7 = 7 - 7 = 0$,
• Denominator: $(5(3) - 7)^{1/3} - 2 = (15 - 7)^{1/3} - 2 = 8^{1/3} - 2 = 2 - 2 = 0$.

Since both the numerator and denominator evaluate to $0$, this is an indeterminate form. We need to simplify the expression.

Step 2: Rationalization

Numerator:

To handle $\sqrt{3x + 40} - 7$, multiply and divide by the conjugate $\sqrt{3x + 40} + 7$: $\sqrt{3x + 40} - 7 = \frac{(\sqrt{3x + 40} - 7)(\sqrt{3x + 40} + 7)}{\sqrt{3x + 40} + 7}.$ The numerator becomes: $(\sqrt{3x + 40})^2 - 7^2 = (3x + 40) - 49 = 3x - 9.$ Thus, the numerator simplifies to: $\frac{3x - 9}{\sqrt{3x + 40} + 7}.$

Denominator:

Rewrite the denominator $(5x - 7)^{1/3} - 2$. Multiply and divide by the conjugate $(5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4$: $(5x - 7)^{1/3} - 2 = \frac{\left((5x - 7)^{1/3} - 2\right)\left((5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4\right)}{(5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4}.$ The numerator becomes: $\left((5x - 7)^{1/3}\right)^3 - 2^3 = (5x - 7) - 8 = 5x - 15.$ Thus, the denominator simplifies to: $\frac{5x - 15}{(5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4}.$

Step 3: Simplify the Overall Expression

The expression becomes: $\frac{\frac{3x - 9}{\sqrt{3x + 40} + 7}}{\frac{5x - 15}{(5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4}}.$ Simplify by multiplying by the reciprocal: $\frac{3x - 9}{\sqrt{3x + 40} + 7} \cdot \frac{(5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4}{5x - 15}.$ Factor $3(x - 3)$ from $3x - 9$ and $5(x - 3)$ from $5x - 15$: $\frac{3(x - 3)}{\sqrt{3x + 40} + 7} \cdot \frac{(5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4}{5(x - 3)}.$ Cancel $x - 3$: $\frac{3}{5} \cdot \frac{(5x - 7)^{2/3} + 2(5x - 7)^{1/3} + 4}{\sqrt{3x + 40} + 7}.$

Step 4: Evaluate the Limit

Substitute $x = 3$ into the simplified expression:

• Numerator: $(5(3) - 7)^{2/3} + 2(5(3) - 7)^{1/3} + 4 = (15 - 7)^{2/3} + 2(15 - 7)^{1/3} + 4 = 8^{2/3} + 2(8^{1/3}) + 4 = 4 + 4 + 4 = 12$,
• Denominator: $\sqrt{3(3) + 40} + 7 = \sqrt{49} + 7 = 7 + 7 = 14$.

Thus, the limit becomes: $\frac{3}{5} \cdot \frac{12}{14} = \frac{3}{5} \cdot \frac{6}{7} = \frac{18}{35}.$

Final Answer:

$\lim_{x \to 3} \frac{\sqrt{3x + 40} - 7}{(5x - 7)^{1/3} - 2} = \frac{18}{35}.$

Would you like a more detailed explanation or clarification?

Follow-up Questions:

1. How is the conjugate method useful in simplifying indeterminate forms?
2. What alternative methods could be used to evaluate limits involving cube roots?
3. Could L'Hôpital's Rule be applied here? Why or why not?
4. What are common indeterminate forms encountered in limits?
5. How does factoring help simplify polynomial expressions in limits?

Tip:

When handling indeterminate forms, rationalizing or factoring are powerful tools to simplify expressions and remove problematic terms.