Limit of n-th roots over e^(sin x) − 1
Evaluate the limit, and determine how its value depends on the positive parameter $n$: $$ L(n) = \lim_{x \to 0^+} \frac{\sqrt[n]{1+x\sin(3x)} \cdot \sqrt[n]{\tan(2x)-1+1}}{e^{\sin x}-1} = \lim_{x \to 0^+} \frac{\sqrt[n]{1+x\sin(3x)} \cdot \sqrt[n]{\,1+\left(\tan(2x)-1\right)\,}}{e^{\sin x}-1}. $$ (Note: the second radical simplifies since $1+\left(\tan(2x)-1\right)=\tan(2x)$.)
Explanation of Method
Each of the three building blocks — the two radicals in the numerator and the exponential in the denominator — is either $1$ plus a small quantity, or reduces directly to a small quantity, as $x\to0^+$. Approximating each with its leading-order behavior (the binomial approximation $\sqrt[n]{1+u}\approx1+\frac{u}{n}$ for the first radical, the direct simplification $1+(\tan2x-1)=\tan2x\approx2x$ for the second, and $e^{\sin x}-1\approx x$ for the denominator) reduces the whole expression to a single power of $x$ times a constant. The sign of that power's exponent, $\frac1n-1$, then determines whether the limit is $0$, a finite constant, or infinite.
- Approximate $1+x\sin(3x)$ and its $n$-th root for small $x$.
- Simplify $1+(\tan(2x)-1)$ to $\tan(2x)$ and approximate its $n$-th root.
- Approximate the denominator $e^{\sin x}-1$.
- Combine everything into a single power of $x$ and analyze the limit based on the sign of the resulting exponent.
Step 1 · Approximate the first radical
As $x\to0^+$, $\sin(3x)\approx3x$, so $x\sin(3x)\approx3x^2\to0$. Using the binomial approximation $\sqrt[n]{1+u}\approx1+\dfrac un$ for small $u$:
$$ \sqrt[n]{1+x\sin(3x)} \approx \sqrt[n]{1+3x^2} \approx 1+\frac{3x^2}{n} \longrightarrow 1 \quad \text{as } x\to0^+. $$
Step 2 · Simplify and approximate the second radical
Since $1+\left(\tan(2x)-1\right) = \tan(2x)$ exactly, and $\tan(2x)\approx2x$ for small $x$:
$$ \sqrt[n]{1+\left(\tan(2x)-1\right)} = \sqrt[n]{\tan(2x)} \approx \sqrt[n]{2x} = (2x)^{1/n}. $$
Step 3 · Approximate the denominator
Since $\sin x \approx x$ for small $x$, and $e^{u}-1\approx u$ for small $u$ (with $u=\sin x$):
$$ e^{\sin x}-1 \approx \sin x \approx x. $$
Step 4 · Combine into a single power of $x$
Putting the pieces together:
$$ \frac{\sqrt[n]{1+x\sin3x}\cdot\sqrt[n]{\tan2x}}{e^{\sin x}-1} \approx \frac{1\cdot(2x)^{1/n}}{x} = 2^{1/n}\, x^{\frac1n-1}. $$
Now examine the sign of the exponent $p=\dfrac1n-1$ as $x\to0^+$:
- If $p>0$ (equivalently $0
- If $p=0$ (equivalently $n=1$): $x^{p}=x^0=1$, so $L(n)=2^{1/n}\cdot1=2^{1}=2$.
- If $p<0$ (equivalently $n>1$): $x^{p}=\dfrac{1}{x^{|p|}}\to+\infty$, so $L(n) = +\infty$.
Summary Table
| Range of $n$ | Exponent $p=\frac1n-1$ | $L(n)$ |
|---|---|---|
$0| $p>0$ | $0$ | |
| $n=1$ | $p=0$ | $2$ |
| $n>1$ | $p<0$ | $+\infty$ |
Additional Notes
- It is worth double-checking the direction of this dependence directly: for $n=2$, the exponent is $p=\frac12-1=-\frac12<0$, and indeed $x^{-1/2}=\frac{1}{\sqrt x}\to+\infty$ as $x\to0^+$, confirming $L(2)=+\infty$ (large $n$ makes the numerator's root "flatter" near $0$ relative to the denominator, so the ratio blows up) — whereas smaller $n$ (below $1$) makes the numerator vanish faster than the denominator, giving $L(n)=0$.
- Since $\tan(2x)$ requires $x\ge0$ near this limit point for the source expression to make direct sense with a real $n$-th root when $n$ is even, this problem is most naturally posed as a one-sided limit $x\to0^+$, as stated.
- The single special value $n=1$ is the borderline case where the numerator's and denominator's rates of vanishing exactly match, producing a finite nonzero limit rather than $0$ or $\infty$ — this kind of "balance point" is typical whenever a limit's outcome is governed by comparing two powers of the same vanishing quantity.
Result
The limit depends on $n$ as follows: