Limit of (fourth root of x − 2)/(sqrt(x) − 4)
Evaluate the limit: $$ L = \lim_{x \to 16} \frac{\sqrt[4]{x} - 2}{\sqrt{x} - 4}. $$
Explanation of Method
Direct substitution of $x=16$ gives $\dfrac{0}{0}$, since $\sqrt[4]{16}=2$ and $\sqrt{16}=4$. Both roots in the expression are powers of the same fourth root, since $\sqrt{x} = \left(\sqrt[4]{x}\right)^2$. Substituting $t=\sqrt[4]{x}$ turns the whole limit into a purely algebraic one in $t$, where the denominator becomes a difference of squares that shares a factor with the numerator.
- Substitute $t = \sqrt[4]{x}$, so that $\sqrt{x} = t^2$ and $x=16$ corresponds to $t=2$.
- Rewrite the limit entirely in terms of $t$.
- Factor the denominator as a difference of squares and cancel the common factor $(t-2)$.
- Substitute $t=2$ into the simplified expression.
Step 1 · Substitute $t=\sqrt[4]{x}$
Let $t = \sqrt[4]{x} = x^{1/4}$, so that $\sqrt{x} = x^{1/2} = \left(x^{1/4}\right)^2 = t^2$. As $x \to 16$, since $16^{1/4}=2$, we have $t \to 2$.
Step 2 · Rewrite the limit in terms of $t$
Substituting into the original expression:
$$ \frac{\sqrt[4]{x}-2}{\sqrt{x}-4} = \frac{t-2}{t^2-4}. $$
So the limit becomes:
$$ L = \lim_{t \to 2} \frac{t-2}{t^2-4}. $$
Step 3 · Factor the denominator and cancel
The denominator is a difference of squares:
$$ t^2 - 4 = (t-2)(t+2). $$
So, for $t \neq 2$:
$$ \frac{t-2}{t^2-4} = \frac{t-2}{(t-2)(t+2)} = \frac{1}{t+2}. $$
Step 4 · Substitute $t=2$
The simplified expression is continuous at $t=2$, so:
$$ L = \lim_{t \to 2} \frac{1}{t+2} = \frac{1}{2+2} = \frac{1}{4}. $$
Summary Table
| Quantity | In terms of $t=\sqrt[4]{x}$ | Value at $t=2$ ($x=16$) |
|---|---|---|
| $\sqrt[4]{x}-2$ | $t-2$ | $0$ |
| $\sqrt{x}-4$ | $t^2-4=(t-2)(t+2)$ | $0$ |
| Simplified ratio | $\dfrac{1}{t+2}$ | $\dfrac{1}{4}$ |
Additional Notes
- Recognizing that $\sqrt{x}$ and $\sqrt[4]{x}$ are both powers of $x^{1/4}$ is the key insight: it converts a limit that mixes two different roots into a single-variable polynomial limit.
- The same problem can be solved with L'Hôpital's Rule (differentiating $x^{1/4}$ and $x^{1/2}$ and substituting $x=16$), which also gives $\frac{1}{4}$; the substitution method here avoids derivatives entirely.
- This technique generalizes: any limit of the form $\lim_{x\to a}\frac{x^{1/m}-a^{1/m}}{x^{1/n}-a^{1/n}}$ where $n$ is a multiple of $m$ can be reduced to a polynomial limit via the substitution $t=x^{1/n}$.
Result
The value of the limit is: