Limit of e^(−x) sin x as x goes to infinity
Evaluate the limit: $$ L = \lim_{x \to +\infty} e^{-x} \sin(x). $$
Explanation of Method
Since $\sin(x)$ oscillates forever and never itself converges, this limit cannot be evaluated by simply taking the limit of each factor. Instead, the bound $-1 \le \sin(x) \le 1$ lets us sandwich $e^{-x}\sin(x)$ between two functions that both tend to the same limit, $0$. The Squeeze Theorem then forces the limit of $e^{-x}\sin(x)$ to be $0$ as well.
- Bound $\sin(x)$ between $-1$ and $1$ for all $x$.
- Multiply the inequality by $e^{-x} > 0$ to bound $e^{-x}\sin(x)$.
- Evaluate the limits of the two bounding functions as $x \to +\infty$.
- Apply the Squeeze Theorem to conclude the limit of the middle expression.
Step 1 · Bound $\sin(x)$
For every real number $x$:
$$ -1 \le \sin(x) \le 1. $$
Step 2 · Multiply through by $e^{-x}$
Since $e^{-x} > 0$ for all $x$, multiplying the inequality by $e^{-x}$ preserves the direction of both inequalities:
$$ -e^{-x} \le e^{-x}\sin(x) \le e^{-x}. $$
Step 3 · Evaluate the bounding limits
As $x \to +\infty$, the exponential $e^{-x} \to 0^+$, so both bounding functions tend to the same value:
$$ \lim_{x \to +\infty} \left(-e^{-x}\right) = 0 \qquad \text{and} \qquad \lim_{x \to +\infty} e^{-x} = 0. $$
Step 4 · Apply the Squeeze Theorem
The function $e^{-x}\sin(x)$ is trapped between two functions that both converge to $0$ as $x \to +\infty$. By the Squeeze Theorem, the trapped function must converge to the same value:
$$ L = \lim_{x \to +\infty} e^{-x}\sin(x) = 0. $$
Summary Table
| Bound | Expression | Limit as $x \to +\infty$ |
|---|---|---|
| Lower | $-e^{-x}$ | $0$ |
| Middle | $e^{-x}\sin(x)$ | $0$ (squeezed) |
| Upper | $e^{-x}$ | $0$ |
Additional Notes
- The Squeeze Theorem is the standard tool whenever a bounded, oscillating factor (like $\sin$ or $\cos$) is multiplied by a factor whose limit is known — the oscillation contributes no obstruction once it is trapped between vanishing bounds.
- Compare this with $\lim_{x\to+\infty} e^{x}\sin(x)$, where the exponential grows instead of decaying: there, the same oscillation is amplified rather than damped, and the limit fails to exist.
- The technique generalizes: for any bounded function $g(x)$ (with $|g(x)| \le M$) and any $h(x) \to 0$, the product $h(x)g(x) \to 0$ as well.
Result
The value of the limit is: