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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.

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    $\begingroup$ $\frac{2x+1}{x-1}>\frac{2x-2}{x-1}=2$ $\endgroup$
    – Empy2
    Commented Aug 15, 2015 at 12:41
  • $\begingroup$ the searched limit is $\infty$ $\endgroup$
    – Dr. Sonnhard Graubner
    Commented Aug 15, 2015 at 12:53
  • $\begingroup$ Is the notation designating the inverse function of the secant function (i.e., $\text{arcsec}(x)=\sec^{-1}(x)$) or the reciprocal of the secant function? I would speculate the former inasmuch as the latter has no limit. $\endgroup$
    – Mark Viola
    Commented Aug 15, 2015 at 16:30

1 Answer 1

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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.

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  • $\begingroup$ 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!! $\endgroup$
    – Mark Viola
    Commented Aug 15, 2015 at 16:28
  • $\begingroup$ @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 $\sec^{-1}$. $\endgroup$
    – Paramanand Singh ♦
    Commented Aug 15, 2015 at 18:36
  • $\begingroup$ 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. $\endgroup$
    – Mark Viola
    Commented Aug 15, 2015 at 18:42
  • $\begingroup$ @ParamanandSingh,$sec^{-1}$ is inverse of sec function only,not the reciprocal of sec function.+1 for ur answer. $\endgroup$
    – learner_avid
    Commented Aug 21, 2015 at 3:47

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