Limit of a square root over a cube root at 8
Evaluate the limit: $$ L = \lim_{x \to 8} \frac{\sqrt{9+2x}-5}{\sqrt[3]{x}-2}. $$
Explanation of Method
Direct substitution of $x=8$ gives $\dfrac{0}{0}$, since $\sqrt{25}-5=0$ and $\sqrt[3]{8}-2=0$. The numerator involves a square root, which can be rationalized by multiplying by its conjugate; the denominator involves a cube root, which can be rationalized using the difference-of-cubes identity $a^3-b^3=(a-b)(a^2+ab+b^2)$. Applying both rationalizations at once produces a common factor of $(x-8)$ that cancels, leaving an expression that is continuous at $x=8$.
- Rationalize the numerator by multiplying by its conjugate $\sqrt{9+2x}+5$.
- Rationalize the denominator by multiplying by $\sqrt[3]{x^2}+2\sqrt[3]{x}+4$, using the difference-of-cubes identity.
- Cancel the common factor of $(x-8)$ that appears in both.
- Substitute $x=8$ into the simplified expression.
Step 1 · Rationalize the numerator
Multiply numerator and denominator by the conjugate $\sqrt{9+2x}+5$:
$$ \left(\sqrt{9+2x}-5\right)\left(\sqrt{9+2x}+5\right) = (9+2x) - 25 = 2x-16 = 2(x-8). $$
Step 2 · Rationalize the denominator
Let $u = \sqrt[3]{x}$, so $u^3 = x$. Using the difference-of-cubes identity $u^3-2^3=(u-2)(u^2+2u+4)$:
$$ x - 8 = u^3 - 8 = (u-2)(u^2+2u+4) = \left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right). $$
So multiplying numerator and denominator by $\sqrt[3]{x^2}+2\sqrt[3]{x}+4$ turns $\sqrt[3]{x}-2$ into $x-8$.
Step 3 · Combine both rationalizations
Multiplying the original fraction by $\dfrac{\sqrt{9+2x}+5}{\sqrt{9+2x}+5}$ and by $\dfrac{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}$ simultaneously:
$$ \frac{\sqrt{9+2x}-5}{\sqrt[3]{x}-2} = \frac{2(x-8)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{(x-8)\left(\sqrt{9+2x}+5\right)}. $$
Cancelling the common factor $(x-8)$ (valid for $x \neq 8$):
$$ \frac{\sqrt{9+2x}-5}{\sqrt[3]{x}-2} = \frac{2\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{\sqrt{9+2x}+5}. $$
Step 4 · Substitute $x=8$
The simplified expression is continuous at $x=8$. Since $\sqrt[3]{8}=2$ and $\sqrt[3]{8^2}=\sqrt[3]{64}=4$:
$$ L = \frac{2(4+2\cdot2+4)}{\sqrt{9+2\cdot8}+5} = \frac{2(4+4+4)}{\sqrt{25}+5} = \frac{2\cdot12}{5+5} = \frac{24}{10} = \frac{12}{5}. $$
Summary Table
| Piece | Rationalized form | Value at $x=8$ |
|---|---|---|
| Numerator $\times$ conjugate | $2(x-8)$ | — |
| Denominator $\times$ cube factor | $(x-8)$ | — |
| Simplified ratio | $\dfrac{2(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{\sqrt{9+2x}+5}$ | $\dfrac{12}{5}$ |
Additional Notes
- Rationalizing a square root uses the conjugate identity $a^2-b^2=(a-b)(a+b)$; rationalizing a cube root uses the difference-of-cubes identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ — both are designed to isolate a linear factor equal to the shared root of both numerator and denominator, here $x-8$.
- The same limit can be found with L'Hôpital's Rule by differentiating $\sqrt{9+2x}$ and $\sqrt[3]{x}$ and substituting $x=8$, which also gives $\frac{12}{5}$ — the rationalization approach avoids derivatives.
- This double-rationalization technique generalizes to any $\frac{0}{0}$ limit built from two different roots of $(x-a)$, by rationalizing each root separately with its matching conjugate identity.
Result
The value of the limit is: