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Evaluate the limit

$$\ell = \lim_{x \rightarrow +\infty} \left( e^{\sqrt{x+2}} + e^{\sqrt{x-2}} - 2 e^{\sqrt{x}} \right)$$

without using differential calculus.

I'm interested in evaluating the above limit using only pure limit theory without using MVT , Taylor etc. The limit is equal to $+\infty$ which is easy to extract using Taylor for example. I'm looking for a way to evaluate it avoiding the big guns.

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    $\begingroup$ what is the point of asking calculus questions that we cannot answer with calculus... $\endgroup$ – zwim Nov 1 '20 at 17:18
  • $\begingroup$ I found the question in a school textbook before MVT or DLH. I'm really stuck. My solution is with MVT. $\endgroup$ – Tolaso Nov 1 '20 at 17:22
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    $\begingroup$ If my calculations (using Taylor expansions) are correct, it looks like the asymptotic behavior is $e^{\sqrt{x+2}} + e^{\sqrt{x-2}} - 2 e^{\sqrt{x}} \sim e^{\sqrt{x}} / x$. $\endgroup$ – Daniel Schepler Nov 1 '20 at 17:57
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    $\begingroup$ Easy to solve with $\lim\limits_{t\to0}(e^t-1-t)/t^2$ known. (But why cut firewood with a jigsaw?..) $\endgroup$ – metamorphy Nov 1 '20 at 19:44
  • $\begingroup$ @metamorphy Care to share a more detailed hint? $\endgroup$ – Tolaso Nov 1 '20 at 19:58
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Let $f(t)=(e^t-1-t)/t^2$; then $e^\sqrt{x+2}+e^\sqrt{x-2}-2e^\sqrt{x}=e^\sqrt{x}g(x)$, where \begin{align*} g(x)&=g_0(x)+g_-(x)+g_+(x), \\g_\pm(x)&=h_\pm^2(x)f\big(h_\pm(x)\big), \\h_\pm(x)&=\sqrt{x\pm 2}-\sqrt{x}=\frac{\pm 2}{\sqrt{x}+\sqrt{x\pm 2}}, \\g_0(x)&=h_+(x)+h_-(x) \\&=\frac{2}{\sqrt{x}+\sqrt{x+2}}-\frac{2}{\sqrt{x}+\sqrt{x-2}} \\&=\frac{2(\sqrt{x-2}-\sqrt{x+2})}{(\sqrt{x}+\sqrt{x+2})(\sqrt{x}+\sqrt{x-2})} \\&=-\frac{8}{(\sqrt{x}+\sqrt{x+2})(\sqrt{x}+\sqrt{x-2})(\sqrt{x+2}+\sqrt{x-2})}. \end{align*} Trivially $\lim\limits_{x\to+\infty}xh_\pm^2(x)=1$ and $\lim\limits_{x\to+\infty}xg_0(x)=0$. Now if we know that $\lim\limits_{t\to 0}f(t)=1/2$, we obtain $\lim\limits_{x\to+\infty}xg(x)=1$, and finally $\lim\limits_{x\to+\infty}e^\sqrt{x}/x=+\infty$ implies that the given limit is $+\infty$.

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