To evaluate the limit

$\lim_{x \to 0} \frac{(x+2)^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x},$

we can use the concept of the derivative or apply L'Hôpital's Rule. This expression resembles the definition of the derivative for a function evaluated at a point. Let’s go through both methods.

Method 1: Recognize as a Derivative

The expression can be seen as the derivative of $f(x) = x^{\frac{1}{3}}$ at $x = 2$: $\lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = f'(2).$ So, let's compute $f'(x)$ for $f(x) = x^{\frac{1}{3}}$.

1. The derivative $f'(x)$ is given by: $f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}.$

2. Now, substituting $x = 2$ into $f'(x)$: $f'(2) = \frac{1}{3 \cdot 2^{\frac{2}{3}}}.$

Thus, the limit evaluates to: $\frac{1}{3 \cdot 2^{\frac{2}{3}}}.$

Method 2: Using L'Hôpital's Rule

Since direct substitution results in an indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule, which states that:

$\lim_{x \to 0} \frac{(x+2)^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x} = \lim_{x \to 0} \frac{\frac{d}{dx} \left( (x+2)^{\frac{1}{3}} \right)}{\frac{d}{dx} \left( x \right)}.$

1. Differentiate the numerator with respect to $x$: $\frac{d}{dx} \left( (x+2)^{\frac{1}{3}} \right) = \frac{1}{3} (x+2)^{-\frac{2}{3}}.$

2. Differentiate the denominator with respect to $x$: $\frac{d}{dx} (x) = 1.$

So the limit becomes: $\lim_{x \to 0} \frac{\frac{1}{3} (x+2)^{-\frac{2}{3}}}{1} = \frac{1}{3 \cdot 2^{\frac{2}{3}}}.$

Final Answer

The evaluated limit is: $\frac{1}{3 \cdot 2^{\frac{2}{3}}}.$

Would you like further details on any step or additional explanation?

Related Questions

1. How can we apply the derivative definition to solve other limit problems?
2. What other forms of indeterminate limits can be solved using L'Hôpital's Rule?
3. How would the solution change if the exponent was different, say $\frac{1}{2}$ instead of $\frac{1}{3}$?
4. What are some other methods to evaluate limits without L'Hôpital's Rule?
5. How does this concept connect to the idea of continuity and differentiability?

Tip

L'Hôpital's Rule is especially useful when you encounter $\frac{0}{0}$ or $\frac{\infty}{\infty}$ indeterminate forms, but remember it requires the derivatives of both the numerator and denominator.