Does e^x sin x have a limit at infinity?
Evaluate the limit (or show that it does not exist): $$ \lim_{x \to +\infty} e^{x} \sin(x). $$
Explanation of Method
Unlike limits that settle to a single number, this one must be tested for existence. A clean way to prove a limit does not exist is to find two sequences $x_n \to +\infty$ along which the function tends to different values (or is unbounded on one and bounded on the other). Here, evaluating $e^x\sin(x)$ at the points where $\sin(x)$ is $0$, $1$, and $-1$ shows the function takes wildly different, unboundedly growing values arbitrarily far out.
- Note that $e^x \to +\infty$ while $\sin(x)$ keeps oscillating between $-1$ and $1$ forever.
- Evaluate the product along specific sequences: $x_n = n\pi$, $x_n = \tfrac{\pi}{2}+2n\pi$, and $x_n = \tfrac{3\pi}{2}+2n\pi$.
- Compare the resulting behaviors to conclude no single limiting value is approached.
Step 1 · Examine the two factors separately
As $x \to +\infty$:
$$ e^x \to +\infty, \qquad -1 \le \sin(x) \le 1 \ \text{ for all } x, \text{ with } \sin(x) \text{ oscillating forever.} $$
Since $\sin(x)$ never settles down, the product $e^x\sin(x)$ is not simply "$\infty$ times a limit" — the oscillating factor must be tracked explicitly.
Step 2 · Evaluate along three sequences
Consider the sequence $x_n = \tfrac{\pi}{2} + 2n\pi$ for $n = 0, 1, 2, \dots$, along which $\sin(x_n) = 1$:
$$ e^{x_n}\sin(x_n) = e^{\pi/2 + 2n\pi} \cdot 1 = e^{\pi/2+2n\pi} \to +\infty \quad \text{as } n \to \infty. $$
Consider the sequence $x_n = \tfrac{3\pi}{2} + 2n\pi$, along which $\sin(x_n) = -1$:
$$ e^{x_n}\sin(x_n) = e^{3\pi/2+2n\pi}\cdot(-1) = -e^{3\pi/2+2n\pi} \to -\infty \quad \text{as } n \to \infty. $$
Consider the sequence $x_n = n\pi$, along which $\sin(x_n) = 0$:
$$ e^{x_n}\sin(x_n) = e^{n\pi} \cdot 0 = 0 \quad \text{for every } n. $$
Step 3 · Conclude that no limit exists
Along three sequences that all tend to $+\infty$, the function $e^x\sin(x)$ tends to $+\infty$, to $-\infty$, and stays exactly at $0$. A limit — finite or infinite in the one-sided sense — would require the function to approach the same value along every such sequence. Since it clearly does not, the limit does not exist.
Summary Table
| Sequence $x_n \to +\infty$ | $\sin(x_n)$ | $e^{x_n}\sin(x_n)$ |
|---|---|---|
| $x_n = n\pi$ | $0$ | $0$ |
| $x_n = \tfrac{\pi}{2}+2n\pi$ | $1$ | $\to +\infty$ |
| $x_n = \tfrac{3\pi}{2}+2n\pi$ | $-1$ | $\to -\infty$ |
Additional Notes
- This is a general technique for disproving the existence of a limit: exhibit two (or more) sequences tending to the same point (or infinity) along which the function tends to different values.
- The behavior here is qualitatively different from $e^{-x}\sin(x)$ (see the related problem where the decaying exponential forces the product to $0$ via the squeeze theorem) — the direction of the exponential's growth is decisive.
- Geometrically, the graph of $e^x\sin(x)$ is a sine wave whose envelope $\pm e^x$ grows without bound, so the oscillations become larger and larger rather than damping out.
Result
The limit does not exist: