Suppose $\lim _{x\to \infty} f(x) = \infty$. Calculate $\lim_{x\rightarrow \infty} \left(\frac{f(x)}{f(x)+1}\right)^{f(x)}$ Suppose $\lim _{x\to \infty} f(x) = \infty$. Calculate: $\lim_{x\rightarrow 
 \infty} \left(\dfrac{f(x)}{f(x)+1}\right)^{f(x)}$
I figured the limit is $\dfrac{1}{e}$, but I have to prove it using the definition of limit, not sure how. Thanks
 A: Notice that $$\left(\frac{f(x)}{f(x)+1} \right)^{f(x)} = \frac{1}{\left(1+\frac{1}{f(x)} \right)^{f(x)}}$$
A: Since $f(x)\to\infty$ as $x\to\infty$, we can think of the limit as one where $n=f(x)$ and $n\to\infty$. Hence, $$\lim_{x\to\infty}\left(\frac{f(x)}{1+f(x)}\right)^{f(x)}=\lim_{n\to\infty}\left(\frac{n}{1+n}\right)^n=\lim_{n\to\infty}\left(\frac{1+n}{n}\right)^{-n}=\left[\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right]^{-1}$$
A: *

*You seem to know that $\lim_{x\to\infty}\left(\frac{x}{1+x}\right)^x=\frac1e$.

*Therefore, you know that for all $\varepsilon>0$ there is $M_\varepsilon$ such that for all $y>M_\varepsilon$, $\left\lvert\frac1e-\left(\frac y{1+y}\right)^y\right\rvert<\varepsilon$.

*You also know by hypothesis that for all $M$ there is $x_M$ such that for all $x>x_M$, $f(x)>M$.

*Finally, for each $\varepsilon>0$ you want $N_\varepsilon$ such that for all $x>N_\varepsilon$, $\left\lvert\frac1e-\left(\frac {f(x)}{1+f(x)}\right)^{f(x)}\right\rvert<\varepsilon$.
A good candidate should be $N_\varepsilon=x_{M_{\varepsilon}}$: for $x>x_{M_\varepsilon}$, $f(x)>M_\varepsilon$ by $(3)$. And, by $(2)$, you have what you want.
A: $$\lim _{ x\rightarrow \infty  } \left( \frac { f(x) }{ f(x)+1 }  \right) ^{ f(x) }=\lim _{ x\rightarrow \infty  } \left( 1-\frac { 1 }{ f(x)+1 }  \right) ^{ f(x) }=\\ =\lim _{ x\rightarrow \infty  }{ \left[ { \left( 1+\frac { -1 }{ f(x)+1 }  \right)  }^{ -\left( f\left( x \right) +1 \right)  } \right]  } ^{ -\frac { f\left( x \right)  }{ 1+f\left( x \right)  }  }={ e }^{ \lim _{ x\rightarrow \infty  }{ \frac { -f\left( x \right) }{ f\left( x \right)+1 }  }  }=\frac { 1 }{ e } $$
A: Set $u=f(x)$. Then $u \to \infty$ as $x \to \infty$, and
$$
\lim_{x \to \infty} \left(\dfrac{f(x)}{f(x)+1}\right)^{f(x)}
= \lim_{u \to \infty} \left(\dfrac{u}{u+1}\right)^{u}
= \lim_{u \to \infty} \left(1-\dfrac{1}{u+1}\right)^{u} \\
= \lim_{u \to \infty} \left(1-\dfrac{1}{u+1}\right)^{u+1} \left(1-\dfrac{1}{u+1}\right)^{-1}
= e^{-1} \cdot 1 = e^{-1}
$$
