Question 1210412: if limit e ^(x) sec (((b sin x))^2) = 0 as x \[LongRightArrow] - \[Infinity] , , then the possible values of b belong to the int erval ....
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! This problem is solved by analyzing the behavior of the two components of the function, $e^x$ and $\sec((b \sin x)^2)$, as $x \to -\infty$.
The limit is given as:
$$\lim_{x \to -\infty} e^x \sec((b \sin x)^2) = 0$$
### 1. Analyze the Exponential Term
As $x \to -\infty$, the term $e^x$ approaches **zero** very rapidly:
$$\lim_{x \to -\infty} e^x = 0$$
### 2. Analyze the Secant Term
The secant function is $\sec(\theta) = \frac{1}{\cos(\theta)}$. The argument is $\theta = (b \sin x)^2$.
* The term $\sin x$ is a bounded function: $-1 \le \sin x \le 1$.
* Therefore, the argument $\theta = (b \sin x)^2$ is also bounded: $0 \le (b \sin x)^2 \le b^2$.
The behavior of the limit is determined by whether the secant term is bounded, or whether it becomes unbounded (oscillating between $-\infty$ and $+\infty$).
$$\sec((b \sin x)^2) = \frac{1}{\cos((b \sin x)^2)}$$
For the limit of the product, $0 \times (\text{Term})$, to be $0$, the $(\text{Term})$ must be **bounded** or, at worst, approach a finite value.
If the secant term is **unbounded**, the limit would be of the indeterminate form $0 \times (\text{Unbounded})$, which may not be $0$. The secant term becomes unbounded (approaches $\pm \infty$) when its denominator, $\cos((b \sin x)^2)$, approaches $0$.
### 3. Determine the Condition for Boundedness
The secant function $\sec(\theta)$ is unbounded if its argument $\theta$ can take the form $\frac{\pi}{2} + n\pi$ for any integer $n$.
For $\sec((b \sin x)^2)$ to be **unbounded**, the value $(b \sin x)^2$ must pass through a singularity, i.e., satisfy:
$$(b \sin x)^2 = \frac{\pi}{2} + n\pi \quad \text{for some integer } n \ge 0$$
(Since the argument is a square, it must be non-negative, so $n$ starts at $0$).
For the secant term to be **bounded** for all $x$, its argument $(b \sin x)^2$ must *never* equal a singularity. Since $0 \le (b \sin x)^2 \le b^2$, we must ensure that the entire range of values $[0, b^2]$ does not contain any singularity.
The first singularity occurs at $\frac{\pi}{2}$.
The first two singularities are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
For the secant term to be bounded, the maximum value of the argument, $b^2$, must be strictly less than the first singularity:
$$b^2 < \frac{\pi}{2}$$
If $b^2 < \frac{\pi}{2}$, then $\frac{1}{\cos((b \sin x)^2)}$ is a continuous function whose argument is always strictly between $0$ and $\frac{\pi}{2}$. Its value is therefore bounded between $\sec(0)=1$ and $\lim_{\theta \to (\pi/2)^-} \sec(\theta) = +\infty$, but the maximum value is $\sec(b^2)$, which is finite.
$$\lim_{x \to -\infty} \underbrace{e^x}_{\to 0} \underbrace{\sec((b \sin x)^2)}_{\text{Bounded}} = 0 \times (\text{Bounded}) = 0$$
### 4. Solve for $b$
We require:
$$b^2 < \frac{\pi}{2}$$
$$|b| < \sqrt{\frac{\pi}{2}}$$
$$-\sqrt{\frac{\pi}{2}} < b < \sqrt{\frac{\pi}{2}}$$
The possible values of $b$ belong to the interval:
$$\left(-\sqrt{\frac{\pi}{2}}, \sqrt{\frac{\pi}{2}}\right)$$
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