Find the following limits: 1. $\lim_{x \to 4} \sqrt[3]{x^3 - 8}$ 2. $\lim_{x \to 0} \frac{(x + 8)^{\frac{1}{3}} - 2}{x}$
Understand the Problem
The question asks to find the limits of two functions. The first one involves a cube root and the second involves a fractional exponent. Both require knowledge of limit evaluation techniques.
Answer
$\lim_{x \to 4} \sqrt[3]{x^3 - 8} = \sqrt[3]{56}$ $\lim_{x \to 0} \frac{(x + 8)^{\frac{1}{3}} - 2}{x} = \frac{1}{12}$
Answer for screen readers
$\lim_{x \to 4} \sqrt[3]{x^3 - 8} = \sqrt[3]{56}$ $\lim_{x \to 0} \frac{(x + 8)^{\frac{1}{3}} - 2}{x} = \frac{1}{12}$
Steps to Solve
- Evaluate the first limit by direct substitution
Since the function $f(x) = \sqrt[3]{x^3 - 8}$ is continuous at $x=4$, we can find the limit by direct substitution: $$ \lim_{x \to 4} \sqrt[3]{x^3 - 8} = \sqrt[3]{4^3 - 8} = \sqrt[3]{64 - 8} = \sqrt[3]{56} $$
- Recognize the second limit as a derivative
The second limit has the form of the definition of a derivative: $$ \lim_{x \to 0} \frac{(x + 8)^{\frac{1}{3}} - 2}{x} $$ This limit can be interpreted as the derivative of the function $f(x) = x^{\frac{1}{3}}$ evaluated at $x=8$. That is, we want to calculate $f'(8)$, where $f(x) = (x+8)^{\frac{1}{3}}$. We can also rewrite it as derivative of the function $g(x) = x^{\frac{1}{3}}$ at $x = 8$, in other words, $g'(8)$.
- Find the derivative
Let's find the derivative of $f(x) = (x+8)^{\frac{1}{3}}$ using the power rule: $$ f'(x) = \frac{1}{3}(x+8)^{\frac{1}{3}-1} = \frac{1}{3}(x+8)^{-\frac{2}{3}} = \frac{1}{3(x+8)^{\frac{2}{3}}} $$
- Evaluate the derivative at $x = 0$ To find the limit, we substitute $x=0$ into the derivative $f'(0)$: $$ f'(0) = \frac{1}{3(0+8)^{\frac{2}{3}}} = \frac{1}{3(8)^{\frac{2}{3}}} = \frac{1}{3(2^3)^{\frac{2}{3}}} = \frac{1}{3(2^2)} = \frac{1}{3(4)} = \frac{1}{12} $$ Alternatively, the limit can be interpreted as the derivative of the function $g(x) = x^{\frac{1}{3}}$ evaluated at $x=8$. That is, we want to calculate $g'(8)$, where $g(x) = x^{\frac{1}{3}}$. $$ g'(x) = \frac{1}{3}x^{-\frac{2}{3}} $$ We substitute $x=8$ into the derivative $g'(8)$: $$ f'(8) = \frac{1}{3}(8)^{-\frac{2}{3}} = \frac{1}{3(8)^{\frac{2}{3}}} = \frac{1}{3(2^3)^{\frac{2}{3}}} = \frac{1}{3(2^2)} = \frac{1}{3(4)} = \frac{1}{12} $$
$\lim_{x \to 4} \sqrt[3]{x^3 - 8} = \sqrt[3]{56}$ $\lim_{x \to 0} \frac{(x + 8)^{\frac{1}{3}} - 2}{x} = \frac{1}{12}$
More Information
The first limit was solved by direct substitution, which is valid since the function is continuous at the point of evaluation. The second limit was recognized as a derivative, allowing us to use differentiation to find the limit.
Tips
A common mistake in evaluating the second limit is failing to recognize it as a derivative. Another common mistake involves incorrectly applying the power rule when finding the derivative or incorrectly simplifying the expression.
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