# On the limit $\lim \limits_{x \rightarrow +\infty} \left( e^{\sqrt{x+2}} + e^{\sqrt{x-2}} - 2 e^{\sqrt{x}} \right)$

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.

• what is the point of asking calculus questions that we cannot answer with calculus... – zwim Nov 1 '20 at 17:18
• I found the question in a school textbook before MVT or DLH. I'm really stuck. My solution is with MVT. – Tolaso Nov 1 '20 at 17:22
• If my calculations (using Taylor expansions) are correct, it looks like the asymptotic behavior is $e^{\sqrt{x+2}} + e^{\sqrt{x-2}} - 2 e^{\sqrt{x}} \sim e^{\sqrt{x}} / x$. – Daniel Schepler Nov 1 '20 at 17:57
• Easy to solve with $\lim\limits_{t\to0}(e^t-1-t)/t^2$ known. (But why cut firewood with a jigsaw?..) – metamorphy Nov 1 '20 at 19:44
• @metamorphy Care to share a more detailed hint? – Tolaso Nov 1 '20 at 19:58

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 $$+\infty$$.