Crazy limit

Take the function $f(x)= e^{\sqrt{x}}$

At $[x^2,x^2+1]$ by mean value theorem we know that $\exists ξ \in [x^2,x^2+1]$ that $f'(ξ)=\frac{f(x^2+1)-f(x^2)}{x^2+1-x^2}$ or $f'(ξ)=e^{\sqrt{x^2+1}}-e^x$ We have : $x\rightarrow \infty \Rightarrow x^2\rightarrow \infty \Rightarrow ξ\rightarrow \infty$ Plus, we have that $f'(ξ)=\frac {e^{\sqrt{ξ}}}{2\sqrt{ξ}}$ so, we evaluate the limit : $\lim_{ξ\to\infty} f'(ξ)$ and by De'l hospital rule we get : $\lim_{x\to\infty} e^{\sqrt{x^2+1}}-e^x = + \infty$

Preliminary results: $\lim_{x \to \infty} \sqrt{x^2 + 1} - x = 0 \\ \lim_{x \to \infty} \frac{x^2 + 1 + x \sqrt{x^2 + 1}}{2x^2} = 1$ These can be proven with basic algebraic manipulations. Keep those in mind when following the computation:

$\lim_{x \to \infty} e^{\sqrt{x^2 - 1}} - e^x = \\ \lim_{x \to \infty} e^x \left( e^{\sqrt{x^2 + 1} - x} - 1 \right) = \lim_{x \to \infty} \frac{e^{\sqrt{x^2 + 1} - x} - 1}{e^{-x}} =$ We may now use L'hopital's rule: $\lim_{x \to \infty} \frac{e^{\sqrt{x^2 + 1} - x} \left( \frac{x}{\sqrt{x^2 + 1}} - 1 \right)}{-e^{-x}} = \\ \lim_{x \to \infty} e^{\sqrt{x^2 + 1}} \left( 1 - \frac{x}{\sqrt{x^2 + 1}} \right) =$

A little algebra before we continue: $1 - \frac{x}{\sqrt{x^2 + 1}} = \frac{\sqrt{x^2 + 1} - x}{\sqrt{x^2 + 1}} = \frac{1}{\sqrt{x^2 + 1}(\sqrt{x^2 + 1} + x)} = \frac{1}{x^2 + 1 + x\sqrt{x^2 + 1}}$ . So the previous limit equals:

$\lim_{x \to \infty} \frac{e^{\sqrt{x^2 + 1}}}{x^2 + 1 + x\sqrt{x^2 + 1}} = \lim_{x \to \infty} \frac{e^x}{2x^2} = \infty$