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I'm a beginner and I'm using a basic graphing calculator. I understand I can input $sec^{(-1)}x$ as $cos^{(-1)}(1/x)$, but even as I'm looking at the graph, I don't get it. How do I determine the answer to be $\pi/2$? Thanks in advance for any help.

Monica
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  • What does $\sec^{-1}(x)$ mean? Is this the multiplicative inverse of $\sec(x)$ or $arcsec(x)$? – Hayden Mar 14 '14 at 01:07
  • @Hayden Based on how the question is written, it can be deduced that $\sec^{-1}(x)=\arcsec(x)$. –  Mar 14 '14 at 01:08
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    Hint: $x$ is getting very large, so $1/x$ is getting very close to $0$ but positive. If my cosine is very close to $0$, roughly what number (between $0$ and $\pi$) am I? – André Nicolas Mar 14 '14 at 01:14
  • @Sanath I assumed as much, but wanted to make sure in case there was a confusion the OP was having which might have helped provide help. – Hayden Mar 14 '14 at 01:20
  • @André Nicolas: Your hint was extremely helpful and helped me determine the answer. – Monica Mar 14 '14 at 01:26
  • Good. It is useful to figure out things oneself. – André Nicolas Mar 14 '14 at 04:53

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$$\sec^{-1}x=\cos^{-1}(1/x)$$ Take the limit: $$\lim_{x\to\infty}\sec^{-1}x=\lim_{x\to\infty}\cos^{-1}(1/x)\to\cos^{-1}(0)=\pi/2\mbox{ in the interval $[0,\pi]$.}$$