Limit of a 31st root minus a 163rd root
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{\sqrt[31]{1-\tan x} - \sqrt[163]{1+\tan x}}{\sin(3x)}. $$
Explanation of Method
Direct substitution gives $\frac{1-1}{0}=\frac00$. Rather than rationalizing (which is impractical with two very different, high-order roots), the binomial approximation $\sqrt[n]{1+u}\approx1+\frac un$ for small $u$ can be applied to each root separately, since both $-\tan x$ and $\tan x$ tend to $0$. This turns both terms in the numerator into simple linear approximations in $x$, whose difference is easy to combine with the denominator's approximation $\sin(3x)\approx3x$.
- Approximate $\tan x$ and $\sin(3x)$ using their small-angle behavior.
- Approximate each root using the binomial approximation $\sqrt[n]{1+u}\approx1+\frac{u}{n}$.
- Combine the two approximated roots to simplify the numerator.
- Divide by the approximated denominator and take the limit.
Step 1 · Small-angle approximations
As $x\to0$:
$$ \tan x \approx x, \qquad \sin(3x) \approx 3x. $$
Step 2 · Approximate each root with the binomial approximation
Using $\sqrt[n]{1+u}\approx1+\dfrac un$ for small $u$, with $u=-\tan x\approx-x$ for the first root and $u=\tan x\approx x$ for the second:
$$ \sqrt[31]{1-\tan x} \approx 1 - \frac{x}{31}, \qquad \sqrt[163]{1+\tan x} \approx 1 + \frac{x}{163}. $$
Step 3 · Combine the two approximated roots
Subtracting:
$$ \sqrt[31]{1-\tan x} - \sqrt[163]{1+\tan x} \approx \left(1-\frac{x}{31}\right) - \left(1+\frac{x}{163}\right) = -\frac{x}{31} - \frac{x}{163} = -x\left(\frac{1}{31}+\frac{1}{163}\right). $$
Combine the fraction using the common denominator $31\times163=5053$:
$$ \frac{1}{31}+\frac{1}{163} = \frac{163+31}{5053} = \frac{194}{5053}. $$
So the numerator is approximately:
$$ \sqrt[31]{1-\tan x} - \sqrt[163]{1+\tan x} \approx -x\cdot\frac{194}{5053}. $$
Step 4 · Divide by the denominator and take the limit
Substituting into the full limit:
$$ L = \lim_{x\to0}\frac{-x\cdot\frac{194}{5053}}{3x} = \frac{-\frac{194}{5053}}{3} = -\frac{194}{5053\times3} = -\frac{194}{15159}. $$
Summary Table
| Term | Binomial approximation | Coefficient of $x$ |
|---|---|---|
| $\sqrt[31]{1-\tan x}$ | $1-\dfrac{x}{31}$ | $-\dfrac1{31}$ |
| $\sqrt[163]{1+\tan x}$ | $1+\dfrac{x}{163}$ | $\dfrac1{163}$ |
| Numerator (difference) | $-\dfrac{194}{5053}x$ | $-\dfrac{194}{5053}$ |
Additional Notes
- The binomial approximation $\sqrt[n]{1+u}\approx1+\frac un$ is simply the first-order Taylor expansion of $(1+u)^{1/n}$ about $u=0$; it applies to each root here separately, avoiding the need to rationalize two roots of very different orders (which would be impractical with indices as large as $31$ and $163$).
- The specific numbers $31$ and $163$ only ever appear through the combination $\frac1{31}+\frac1{163}=\frac{194}{5053}$ — this is a good illustration that the exact indices of the roots matter only through this single combined coefficient in the final answer.
- This same limit could be found with L'Hôpital's Rule, differentiating each root term using the chain rule and $\frac{d}{dx}\tan x=\sec^2x\to1$ at $x=0$, which reproduces the same coefficients $\frac1{31}$ and $\frac1{163}$ — the binomial approach here is simply a faster route to the same linear-order information.
Result
The value of the limit is: