Limit of e^(a^x) / x^k as x approaches 0
Given $$ \lim_{x \to 0^+} \frac{e^{a^x}}{x^k} = L, $$ where $k>0$ and $a>0$, find $L$ for the different possible values of $a$.
Explanation of Method
Before reaching for series expansions or L'Hôpital's Rule, it always pays to check whether the numerator itself has a finite, nonzero limit. Here, $a^x$ is continuous in $x$ for any fixed $a>0$, so $a^x \to a^0 = 1$ as $x\to0^+$, regardless of which positive value $a$ takes. That means the numerator $e^{a^x}$ tends to the single fixed constant $e^1=e$ — it does not shrink toward $0$ at all. Since the denominator $x^k$ does shrink to $0^+$ (because $k>0$), the ratio of a nonzero constant over something vanishing must blow up, no matter the precise value of $a$ or $k$.
- Evaluate $\lim_{x\to0^+} a^x$ using continuity of $a^x$ in $x$.
- Evaluate the resulting limit of the numerator $e^{a^x}$.
- Compare this nonzero numerator limit against the vanishing denominator $x^k$ to conclude the overall limit.
Step 1 · The exponent $a^x$ tends to $1$
For any fixed $a>0$, the function $x \mapsto a^x = e^{x\ln a}$ is continuous. So as $x \to 0^+$:
$$ a^x \to a^0 = 1 \qquad \text{for every } a > 0. $$
Step 2 · The numerator tends to the constant $e$
Since $e^{(\cdot)}$ is continuous, composing with the result of Step 1 gives:
$$ \lim_{x\to0^+} e^{a^x} = e^{\lim_{x\to0^+}a^x} = e^{1} = e. $$
Crucially, this holds for every $a>0$ — the numerator always settles down to the same fixed, nonzero constant $e$, with no dependence on $a$ surviving in the limit.
Step 3 · Compare against the vanishing denominator
Since $k>0$, the denominator satisfies $x^k \to 0^+$ as $x\to0^+$. A ratio whose numerator tends to a fixed positive constant while its denominator tends to $0^+$ must grow without bound:
$$ L = \lim_{x\to0^+}\frac{e^{a^x}}{x^k} = \frac{e}{0^+} = +\infty, \qquad \text{for every } a>0 \text{ and every } k>0. $$
The same conclusion holds at the boundary case $a=1$ (where $a^x=1$ identically, so the numerator is exactly $e$ for all $x$) and, by a similar direct check, at $a=0$ (where $0^x=0$ for $x>0$, so the numerator is exactly $e^0=1$, still a nonzero constant). For $a<0$, $a^x$ is not real-valued for all real $x$ near $0$, so the limit is not defined in the real numbers.
Summary Table
| Value of $a$ | $\lim_{x\to0^+} e^{a^x}$ | $L=\lim_{x\to0^+} e^{a^x}/x^k$ |
|---|---|---|
| $a>0,\ a\neq1$ | $e$ | $+\infty$ |
| $a=1$ | $e$ | $+\infty$ |
| $a=0$ | $1$ | $+\infty$ |
| $a<0$ | — | not defined (real) |
Additional Notes
- A common error (worth flagging explicitly) is to expand $e^{a^x}$ with a Taylor-type approximation such as $e^{a^x}\approx e(1+x\ln a)$ and then divide by $x^k$ term by term, concluding that at $k=1$ the leading term $e/x^k$ somehow disappears, leaving only a finite correction like $e(1+\ln a)$. That is incorrect: the leading term $e/x^k$ never vanishes — as $x\to0^+$ it is exactly what forces the whole expression to diverge, at every value of $k>0$, including $k=1$.
- The general principle illustrated here is worth remembering as a first check on any limit of a quotient: if the numerator tends to a finite nonzero value and the denominator tends to $0$, the limit is automatically $\pm\infty$ (or does not exist, if the denominator can vanish from both signs) — no further expansion is needed.
- Had the exponent in the numerator instead been something that itself tends to $0$ (for instance $e^{a^x}-e$, or $e^{a^x - 1}$ type expressions), the numerator would vanish too, and a genuine $\frac00$-type analysis with series expansions would become necessary.
Result
For every allowed value of $a>0$ (and $a=0$) and every $k>0$: