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

clearly this is of the form $1^\infty$ so we can use short cut method to write...$$e^{\lim_{ x\to\infty}{(x+2)(\frac{x-1}{x+1}-1)}}$$ After this clearly the answer is $e^{-2}$ (as degree of numerator and denominator are same so only -2 remains). I want to know other ways to solve this (may be using some standard formulae).

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  • $\begingroup$ I would say that what you have done is the standard technique. That, or what is mathematically the same... $L = \lim f(x) \implies \ln L = \lim \ln (f(x))$ $\endgroup$
    – Doug M
    Commented Aug 29, 2016 at 16:40

2 Answers 2

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Note that we can write

$$\begin{align} \lim_{x\to \infty}\left(\frac{x-1}{x+1}\right)^{x+2}&=\lim_{x\to \infty}\left(1-\frac{2}{x+1}\right)^{x+2}\\\\ &=\lim_{x\to \infty}\left(1-\frac{2}{x+1}\right)^{x+1}\left(1-\frac{2}{x+1}\right)\\\\ &=\lim_{x\to \infty}\left(1-\frac{2}{x+1}\right)^{x+1}\,\lim_{x\to \infty}\left(1-\frac{2}{x+1}\right)\\\\ &=e^{-2} \end{align}$$

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  • $\begingroup$ very clean way to show the intuitive $\endgroup$
    – operatorerror
    Commented Aug 29, 2016 at 18:01
  • $\begingroup$ @qbert Thank you! Much appreciative. $\endgroup$
    – Mark Viola
    Commented Aug 29, 2016 at 18:05
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$$\lim_{x\to \infty}\left(\dfrac{x-1}{x+1}\right)^{x+2}= \lim_{x\to \infty}\left(1-\dfrac2{x+1}\right)^{x+2}$$ $$=\lim_{x\to \infty}\left(\left(1-\dfrac2{x+1}\right)^\dfrac{x+1}2\right)^\dfrac{2(x+2)}{x+1}= e^{-2}.$$ Used formula $$\lim_{t\to \infty}\left(1-\dfrac1{t}\right)^t=e^{-1}.$$

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