Limit of $\mathrm{e}^{\sqrt{x+1}} - \mathrm{e}^{\sqrt{x}}$ How can I calculate the below limit? 
$$
\lim\limits_{x\to \infty} \left( \mathrm{e}^{\sqrt{x+1}} - \mathrm{e}^{\sqrt{x}} \right)
$$
In fact I know should use the L’Hospital’s Rule, but I do not how to use it.
 A: One may write, as $x \to \infty$,
$$
\begin{align}
e^{\sqrt{x+1}} - e^{\sqrt{x}}&=e^{\sqrt{x}}\left(e^{\sqrt{x+1}-\sqrt{x}} - 1\right)
\\\\&=e^{\sqrt{x}}\left(e^{\frac1{\sqrt{x+1}+\sqrt{x}}} - 1\right)
\\\\&=e^{\sqrt{x}}\left(1+\frac1{\sqrt{x+1}+\sqrt{x}}+O\left(\frac1x\right) - 1\right)
\\\\& \sim \frac{e^{\sqrt{x}}}{2\sqrt{x}}
\end{align}
$$ which goes to $\infty$.
A: Using the fact that $\lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x }  } =1\\ \\ $ we can write 
$$\lim _{ x\rightarrow \infty  }{ \left( e^{ \sqrt { x+1 }  }-e^{ \sqrt { x }  } \right)  } =\lim _{ x\rightarrow \infty  }{ { e }^{ \sqrt { x }  }\left( e^{ \sqrt { x+1 } -\sqrt { x }  }-1 \right)  } =\lim _{ x\rightarrow \infty  }{ { e }^{ \sqrt { x }  }\left( \frac { e^{ \frac { 1 }{ \sqrt { x+1 } +\sqrt { x }  }  }-1 }{ \frac { 1 }{ \sqrt { x+1 } +\sqrt { x }  }  }  \right)  } \cdot \frac { 1 }{ \sqrt { x+1 } +\sqrt { x }  } =\\ =\lim _{ x\rightarrow \infty  }{ \frac { { e }^{ \sqrt { x }  } }{ \sqrt { x+1 } +\sqrt { x }  }  } =+\infty $$
