Evaluate the following limit: $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x^{1/3} - 2}$
Understand the Problem
The question asks us to evaluate the limit of a given function as x approaches 8. The function involves a cube root and a fractional exponent. Specifically, we need to find (\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x^{1/3} - 2}).
Answer
If the question was $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x^{1/3} - 2}$, the answer is $1$. If the question intended to be $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}$, the answer is $\frac{1}{12}$.
Answer for screen readers
If the question was $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x^{1/3} - 2}$, the answer is $1$.
If the question intended to be $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}$, the answer is $\frac{1}{12}$.
Steps to Solve
- Recognize the indeterminate form
If we directly substitute $x = 8$ into the expression, we get $\frac{\sqrt[3]{8} - 2}{8^{1/3} - 2} = \frac{2 - 2}{2 - 2} = \frac{0}{0}$, which is an indeterminate form. This means we can use L'Hôpital's Rule or algebraic manipulation to find the limit.
- Apply L'Hôpital's Rule
L'Hôpital's Rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the limit exists.
Let $f(x) = \sqrt[3]{x} - 2 = x^{1/3} - 2$ and $g(x) = x^{1/3} - 2$. Then, $f'(x) = \frac{1}{3}x^{-2/3}$ and $g'(x) = \frac{1}{3}x^{-2/3}$.
Therefore, $\lim_{x \to 8} \frac{f'(x)}{g'(x)} = \lim_{x \to 8} \frac{\frac{1}{3}x^{-2/3}}{\frac{1}{3}x^{-2/3}} = \lim_{x \to 8} 1 = 1$
- Revised Application of L'Hopital's Rule Let's assume there was an error in copying the question, and the problem was instead $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}$
Then $f(x) = \sqrt[3]{x} - 2 = x^{1/3} - 2$ and $g(x) = x - 8$.
Then, $f'(x) = \frac{1}{3}x^{-2/3}$ and $g'(x) = 1$.
Applying L'Hopital's rule: $\lim_{x \to 8} \frac{f'(x)}{g'(x)} = \lim_{x \to 8} \frac{\frac{1}{3}x^{-2/3}}{1} = \frac{1}{3}(8)^{-2/3} = \frac{1}{3} (2^3)^{-2/3} = \frac{1}{3} (2^{-2}) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$
If the question was $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x^{1/3} - 2}$, the answer is $1$.
If the question intended to be $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}$, the answer is $\frac{1}{12}$.
More Information
L'Hôpital's Rule is a powerful tool for evaluating limits of indeterminate forms. It involves taking the derivative of the numerator and the derivative of the denominator separately, and then evaluating the limit of the new fraction.
Tips
A common mistake is to try to directly substitute the value that $x$ approaches into the expression without checking for indeterminate forms. This can lead to incorrect results. Also, students sometime incorrectly compute the derivatives. Additionally, students may forget to apply the chain rule for the derivatives.
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