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$\lim_{x\to\infty}\sec^{-1}\left(\frac{2x+1}{x-1}\right)^x$

$\lim_{x\to\infty}\sec^{-1}\left(\frac{2x+1}{x-1}\right)^x=\sec^{-1}\lim_{x\to\infty}\left(\frac{2x+1}{x-1}\right)^x$

Let $L=\lim_{x\to\infty}\left(\frac{2x+1}{x-1}\right)^x$

$\Rightarrow \log L=\lim_{x\to\infty}x\log\left(\frac{2x+1}{x-1}\right)$
and then i stuck.Help me.

Paramanand Singh
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learner_avid
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1 Answers1

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Clearly when $x \to \infty$ the fraction $(2x + 1)/(x - 1) \to 2$ and hence $\left(\dfrac{2x + 1}{x - 1}\right)^{x} \to \infty$. Now it is easy to see that $\sec^{-1}(y) = \cos^{-1}(1/y)$ and hence $$\sec^{-1}\left(\dfrac{2x + 1}{x - 1}\right)^{x} = \cos^{-1}\left\{\dfrac{1}{\left(\dfrac{2x + 1}{x - 1}\right)^{x}}\right\} \to \cos^{-1}(0) = \frac{\pi}{2}$$ Hence the answer is $\pi/2$. Its rather surprising that OP has accepted the wrong answer.

Paramanand Singh
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  • I suppose the interpretation of $\sec^{-1}(x)$ as $\cos (x)$ prevailed. That is why the notation $arcsec(x)$ is less ambiguous. So, ... a +1 for you correct result!! – Mark Viola Aug 15 '15 at 16:28
  • @Dr.MV: Yes the notation "arc..." for inverse trigonometric functions is less ambiguous. But in India this notation is virtually non-existent. I got to know about this notation in my college years when studying books by foreign authors. In fact in Indian books there is special emphasis given (during intro to inverse trig functions) that notation $\sin^{-1}$ is not reciprocal of sine but rather its inverse function. Anyway I hope OP replies to your comment to clarify what he meant by $\sec^{-1}$. – Paramanand Singh Aug 15 '15 at 18:36
  • Yes, the $f^{-1}(x)$ is still prevalent in the literature. I understand the rationale for representing inverse operations. However, in some cases one cannot distinguish what type of inversion that notation designates. For an operator $A$, it is usually clear that $A^{-1}$ denotes the inverse. But with the trigs, it simply is not. I believe that your answer is the one of interest here since the other interpretation just leaves an undefined limit. – Mark Viola Aug 15 '15 at 18:42
  • @ParamanandSingh,$sec^{-1}$ is inverse of sec function only,not the reciprocal of sec function.+1 for ur answer. – learner_avid Aug 21 '15 at 03:47