Limit of (1 + fifth root of x) at x = −1
Evaluate the limit: $$ L = \lim_{x \to -1} \frac{1+\sqrt[5]{x}}{1+\sqrt[3]{x}}. $$
Explanation of Method
Direct substitution gives $\frac{1+(-1)}{1+(-1)}=\frac00$. The two roots involved, a fifth root and a cube root, are both odd roots — well-defined for negative numbers — and both are powers of a common fifteenth root, since $15$ is a multiple of both $5$ and $3$. Substituting $t=\sqrt[15]{x}$ turns both roots into integer powers of $t$, converting the whole limit into a ratio of polynomials in $t$ that can be factored and simplified directly, with no risk of the sign errors that fractional-exponent shortcuts can introduce for negative bases.
- Substitute $t=\sqrt[15]{x}$, so that $\sqrt[5]{x}=t^3$ and $\sqrt[3]{x}=t^5$.
- Rewrite the limit as a ratio of polynomials in $t$, with $t\to-1$.
- Factor both $1+t^3$ and $1+t^5$ using the sum-of-powers identities, and cancel the shared factor $(1+t)$.
- Substitute $t=-1$ into the simplified ratio.
Step 1 · Substitute $t=\sqrt[15]{x}$
Since $15$ is a common multiple of $5$ and $3$, let $t=x^{1/15}$ (the real fifteenth root, well-defined for negative $x$ since $15$ is odd). Then:
$$ \sqrt[5]{x} = x^{1/5} = \left(x^{1/15}\right)^3 = t^3, \qquad \sqrt[3]{x} = x^{1/3} = \left(x^{1/15}\right)^5 = t^5. $$
As $x \to -1$, since $(-1)^{1/15}=-1$ (the real fifteenth root of $-1$), we have $t \to -1$.
Step 2 · Rewrite the limit in terms of $t$
Substituting:
$$ L = \lim_{t \to -1} \frac{1+t^3}{1+t^5}. $$
Step 3 · Factor and cancel the shared root
Using the sum-of-cubes and sum-of-fifth-powers identities:
$$ 1+t^3 = (1+t)\left(1-t+t^2\right), \qquad 1+t^5 = (1+t)\left(1-t+t^2-t^3+t^4\right). $$
For $t \neq -1$, the common factor $(1+t)$ cancels:
$$ \frac{1+t^3}{1+t^5} = \frac{1-t+t^2}{1-t+t^2-t^3+t^4}. $$
Step 4 · Substitute $t=-1$
The simplified expression is a ratio of polynomials, continuous at $t=-1$. Evaluating the numerator:
$$ 1-(-1)+(-1)^2 = 1+1+1 = 3. $$
Evaluating the denominator:
$$ 1-(-1)+(-1)^2-(-1)^3+(-1)^4 = 1+1+1+1+1 = 5. $$
So:
$$ L = \frac{3}{5}. $$
Summary Table
| Expression | Factored form | Value at $t=-1$ (reduced) |
|---|---|---|
| $1+t^3$ | $(1+t)(1-t+t^2)$ | $3$ |
| $1+t^5$ | $(1+t)(1-t+t^2-t^3+t^4)$ | $5$ |
Additional Notes
- Working through a common substitution $t=x^{1/15}$ avoids a subtle pitfall of applying L'Hôpital's Rule directly with fractional exponents on a negative base: combining $x^{-4/5}$ and $x^{-2/3}$ into a single power $x^{-2/15}$ and then evaluating $(-1)^{-2/15}$ carelessly can produce a sign error, since $(-1)^{-2/15}=\left((-1)^{-2}\right)^{1/15}=1^{1/15}=1$, not $-1$ — the sign must be tracked through the odd-root branch consistently, which the polynomial substitution here does automatically.
- This can be double-checked with L'Hôpital's Rule applied carefully: differentiating $1+x^{1/5}$ and $1+x^{1/3}$ termwise and evaluating each fractional power at $x=-1$ via its real odd-root branch (rather than combining exponents) gives $\frac{(1/5)\cdot1}{(1/3)\cdot1}=\frac35$, confirming the same answer.
- The general technique — substituting the least common root to clear all the fractional exponents at once — works for any limit built from several different roots of the same base, converting it into an ordinary polynomial limit that can be handled by factoring.
Result
The value of the limit is: