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let be the function $ ( \epsilon x +1)^{1/\epsilon}=f(x) $, i know that in the limit

$ \epsilon \to 0 $ then $f(x)=e^{x}$ , however i would like to know what happens if i have the function

$ ( \epsilon x +a)^{1/\epsilon}=f(x) $, in the cases $ a >1$ , $a<1$

in the first case if $a>1 $ and 'x' is positive the function is $f(x)=\infty$ , for the other case with $ a<1$ i believe that $ f(x)=0$ , but i am not sure.

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up vote 2 down vote accepted

A simple way to do this is this one. $$ f_a(x) = (\varepsilon x + a)^{1/\varepsilon} = a^{1/\varepsilon} \cdot (\varepsilon(x/a) + 1)^{1/\varepsilon}. $$ For the case $a = 1$, you know that the limit exists and no factor $a^{1/\varepsilon}$ appears. It gives you $f_1(x) = e^x$. Now this show that for $a > 0$, $$ f_a(x) = \left( \lim_{\varepsilon \to 0} \, a^{1/\varepsilon} \right) e^{x/a} = \left( \lim_{y \to \infty} \, a^y \right) e^{x/a}. $$ Since the limit in $y$ is well known to be $0$ if $0 < a < 1$ and $\infty$ if $a > 1$, you have your desired result.

Hope that helps,

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You are correct. For a constant $a\in \mathbb{R}^+$ such that $a<1$, we have $\lim_{\epsilon\rightarrow 0} ( \epsilon x +a)^{1/\epsilon} = 0,\forall x\in\mathbb{R}$ because this is trivially true for $x=0$ and any $0\neq x\in \mathbb{R}$ we have that $$\epsilon < \frac{1-a}{2|x|}\implies 0 < ( \epsilon x +a)^{1/\epsilon} < ( a + \frac{1-a}{2})^{1/\epsilon}$$ and since $a + \frac{1-a}{2} < 1$ we have $\lim_{\epsilon\rightarrow 0}( a + \frac{1-a}{2})^{1/\epsilon} = 0$ so the desired result follows from squeeze theorem, as we can neglect all $\epsilon \geq \frac{1-a}{2|x|}$ when we take the limit.

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