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Evaluate $\lim\limits_{x\to+\infty} \dfrac{x-\sqrt{e^x-1}}{x+\sqrt{e^x+2}}$

I think the limit clearly evaluates to $-1$ as $\sqrt{e^x-1}$ and $\sqrt{e^x+2}$ dominate the value for the numerator and the denominator respectively. However, I am unable to express this idea rigorously. I am also looking for other ways (e.g., L'Hopital's rule) to compute this limit. I've tried multiplying the limit by it's conjugate but to no avail. Thanks in advance.

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    $\begingroup$ Hint: multiply the numerator & denominator by $e^{-x/2}$. $\endgroup$
    – J.G.
    Commented Dec 6, 2022 at 8:39
  • $\begingroup$ Oh I see, thank you $\endgroup$
    – Y.T.
    Commented Dec 6, 2022 at 8:43
  • $\begingroup$ @J.G. I am confused. Intuitively, I (I'm not the OP) know that the answer is $-1$. However, when I apply L'Hopital's rule I get $$\frac{1 - \frac{1}{2\sqrt{e^x -1}}}{1 + \frac{1}{2\sqrt{e^x +2}}}.$$ As $x \to \infty,$ both the numerator and denominator go to $1$, so I get the result of $1$ rather than $-1$. What mistake am I making? $\endgroup$
    – user2661923
    Commented Dec 6, 2022 at 9:06
  • $\begingroup$ @user2661923 $\frac{d}{dx}\sqrt{e^x+c}=\frac{\color{red}{e^x}}{2\sqrt{e^x+c}}$, so your use of L'Hôpital's rule gives a numerator and denominator each asymptotic to a multiple of $e^{x/2}$, and that factor should still be cancelled. $\endgroup$
    – J.G.
    Commented Dec 6, 2022 at 9:22
  • $\begingroup$ @J.G. Thanks, my blind spot. $\endgroup$
    – user2661923
    Commented Dec 6, 2022 at 9:23

1 Answer 1

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Here is another method (probably fastest way), using asymptotic behaviour. Recall that :

$\displaystyle \lim_{x\to +\infty}\frac{x}{e^x}=0$

Hence we have that:

$$\frac{x-\sqrt{e^x-1}}{x+\sqrt{e^x+2}}\sim \frac{-\sqrt{e^x-1}}{\sqrt{e^x+2}}\sim \frac{-\sqrt{e^x}}{\sqrt{e^x}}\to -1 \quad \text{as} \to +\infty $$

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