Limit of x² / e^x as x goes to infinity
Evaluate the limit: $$ L = \lim_{x \to +\infty} \frac{x^2}{e^x}. $$
Explanation of Method
As $x\to+\infty$, both $x^2\to\infty$ and $e^x\to\infty$, giving the indeterminate form $\dfrac\infty\infty$. Since the numerator is a polynomial of degree $2$, differentiating twice reduces it all the way down to a constant, while each differentiation leaves the denominator unchanged as $e^x$ (since $e^x$ is its own derivative). This makes repeated L'Hôpital's Rule a natural fit: two applications are exactly enough to eliminate the polynomial entirely.
- Apply L'Hôpital's Rule once, differentiating $x^2$ and $e^x$.
- Recognize the new limit is still $\frac\infty\infty$, and apply L'Hôpital's Rule a second time.
- Evaluate the resulting limit, now with a constant numerator.
Step 1 · Apply L'Hôpital's Rule once
Since both $x^2\to\infty$ and $e^x\to\infty$ as $x\to+\infty$, L'Hôpital's Rule applies:
$$ \lim_{x\to+\infty}\frac{x^2}{e^x} = \lim_{x\to+\infty}\frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}(e^x)} = \lim_{x\to+\infty}\frac{2x}{e^x}. $$
Step 2 · Apply L'Hôpital's Rule a second time
The new limit $\dfrac{2x}{e^x}$ still has the form $\dfrac\infty\infty$ (since $2x\to\infty$ and $e^x\to\infty$), so apply L'Hôpital's Rule again:
$$ \lim_{x\to+\infty}\frac{2x}{e^x} = \lim_{x\to+\infty}\frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(e^x)} = \lim_{x\to+\infty}\frac{2}{e^x}. $$
Step 3 · Evaluate the final limit
Since $e^x\to+\infty$ as $x\to+\infty$, the constant numerator $2$ divided by an unboundedly growing denominator tends to $0$:
$$ L = \lim_{x\to+\infty}\frac{2}{e^x} = 0. $$
Summary Table
| Application | Resulting expression | Form |
|---|---|---|
| Original | $\dfrac{x^2}{e^x}$ | $\frac\infty\infty$ |
| After 1st L'Hôpital | $\dfrac{2x}{e^x}$ | $\frac\infty\infty$ |
| After 2nd L'Hôpital | $\dfrac{2}{e^x}$ | $\to0$ |
Additional Notes
- Every time L'Hôpital's Rule is applied to a polynomial-over-exponential ratio, the polynomial's degree drops by one while the exponential stays exactly $e^x$ — so a degree-$n$ polynomial over $e^x$ always resolves after exactly $n$ applications of the rule, ending in a constant over $e^x\to0$.
- This confirms the general principle that exponential functions grow faster than any polynomial: for every fixed power $n$, $\lim_{x\to+\infty}\dfrac{x^n}{e^x}=0$, which can be proven by the same repeated-differentiation argument used here.
- Each time the rule is reapplied, it is important to first re-check that the new ratio is still an indeterminate form ($\frac00$ or $\frac\infty\infty$) — here both intermediate ratios were confirmed to be $\frac\infty\infty$ before differentiating again.
Result
The value of the limit is: