Q.Evaluate $\lim_{x \to 0} \dfrac{(x + 2)^{\frac{1}{3}} - 2^{\frac{1}{3}}}{x}$.

Concept: Derivative from first principles (limit definition).

The expression has the form $\lim_{x \to 0} \dfrac{f(x) - f(0)}{x - 0}$ where $f(x) = (x+2)^{1/3}$.

This is precisely the derivative $f'(0)$ by definition.

Compute $f'(x)$ using the power rule:

Evaluate at $x = 0$:

$f'(0) = \frac{1}{3}(2)^{-2/3} = \frac{1}{3 \cdot 2^{2/3}} = \frac{1}{3 \cdot \sqrt[3]{4}}$ …

This limit has the classic $\frac{0}{0}$ indeterminate form when you substitute $x = 0$ directly. The structure—a difference quotient—tells us exactly what's happening: we're measuring the instantaneous rate of change of the cube-root function at a specific point.

If you set $f(t) = t^{1/3}$, then the limit becomes

which is precisely the definition of $f'(2)$, the derivative of $f$ at $t = 2$. This is the conceptual heart of the problem: the limit is asking for a derivative in disguise.

Solution

1. Identify the function and point Let $f(t) = t^{1/3}$. The given limit rewrites as

2. Compute the derivative Using the power rule, $f(t) = t^{1/3}$ gives

3. Evaluate at $t = 2$ Substitute $t = 2$ into the derivative:

4. Simplify (optional)

   You can rationalize if needed. Multiply numerator and denominator by $2^{1/3}$:

   $\frac{1}{3 \cdot 2^{2/3}} \cdot \frac{2^{1/3}}{2^{1/3}} = \frac{2^{1/3}}{3 \cdot 2} = \frac{2^{1/3}}{6} = \frac{\sqrt[3]{2}}{6}$ …