document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210412>Question 1210412</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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 .... </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852317 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852317\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This problem is solved by analyzing the behavior of the two components of the function, $e^x$ and $\sec((b \sin x)^2)$, as $x \to -\infty$.\r<BR>\n" );
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document.write( "The limit is given as:<BR>\n" );
document.write( "$$\lim_{x \to -\infty} e^x \sec((b \sin x)^2) = 0$$\r<BR>\n" );
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document.write( "### 1. Analyze the Exponential Term<BR>\n" );
document.write( "As $x \to -\infty$, the term $e^x$ approaches **zero** very rapidly:<BR>\n" );
document.write( "$$\lim_{x \to -\infty} e^x = 0$$\r<BR>\n" );
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document.write( "### 2. Analyze the Secant Term<BR>\n" );
document.write( "The secant function is $\sec(\theta) = \frac{1}{\cos(\theta)}$. The argument is $\theta = (b \sin x)^2$.\r<BR>\n" );
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document.write( "* The term $\sin x$ is a bounded function: $-1 \le \sin x \le 1$.<BR>\n" );
document.write( "* Therefore, the argument $\theta = (b \sin x)^2$ is also bounded: $0 \le (b \sin x)^2 \le b^2$.\r<BR>\n" );
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document.write( "The behavior of the limit is determined by whether the secant term is bounded, or whether it becomes unbounded (oscillating between $-\infty$ and $+\infty$).\r<BR>\n" );
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document.write( "$$\sec((b \sin x)^2) = \frac{1}{\cos((b \sin x)^2)}$$\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "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 $0$.\r<BR>\n" );
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document.write( "### 3. Determine the Condition for Boundedness\r<BR>\n" );
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document.write( "The secant function $\sec(\theta)$ is unbounded if its argument $\theta$ can take the form $\frac{\pi}{2} + n\pi$ for any integer $n$.\r<BR>\n" );
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document.write( "For $\sec((b \sin x)^2)$ to be **unbounded**, the value $(b \sin x)^2$ must pass through a singularity, i.e., satisfy:<BR>\n" );
document.write( "$$(b \sin x)^2 = \frac{\pi}{2} + n\pi \quad \text{for some integer } n \ge 0$$<BR>\n" );
document.write( "(Since the argument is a square, it must be non-negative, so $n$ starts at $0$).\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "The first singularity occurs at $\frac{\pi}{2}$.<BR>\n" );
document.write( "The first two singularities are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.\r<BR>\n" );
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document.write( "For the secant term to be bounded, the maximum value of the argument, $b^2$, must be strictly less than the first singularity:<BR>\n" );
document.write( "$$b^2 < \frac{\pi}{2}$$\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "$$\lim_{x \to -\infty} \underbrace{e^x}_{\to 0} \underbrace{\sec((b \sin x)^2)}_{\text{Bounded}} = 0 \times (\text{Bounded}) = 0$$\r<BR>\n" );
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document.write( "### 4. Solve for $b$<BR>\n" );
document.write( "We require:<BR>\n" );
document.write( "$$b^2 < \frac{\pi}{2}$$<BR>\n" );
document.write( "$$|b| < \sqrt{\frac{\pi}{2}}$$<BR>\n" );
document.write( "$$-\sqrt{\frac{\pi}{2}} < b < \sqrt{\frac{\pi}{2}}$$\r<BR>\n" );
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document.write( "The possible values of $b$ belong to the interval:<BR>\n" );
document.write( "$$\left(-\sqrt{\frac{\pi}{2}}, \sqrt{\frac{\pi}{2}}\right)$$ </SPAN>\n" );
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