Prove the limit of $\lim_{x \to 0^+}e^{\frac{1}{x}}$ and $\lim_{x \to 0^-}e^{\frac{1}{x}}$ I have been given the following task:
Determine whether a limit exists and if so determine it. I need to analyze the following two functions.
$$(1) \lim_{x \to 0^+}e^{\frac{1}{x}}$$
$$(2) \lim_{x \to 0^-}e^{\frac{1}{x}}$$
I know for a fact that a limit for the first function does not exist because for very small positiv values the term $\frac{1}{x}$ approaches infinity.
For the second function a limit does exist. As x approaches very small negative values the term $\frac{1}{x}$ approaches negative infinity. So it follows that $\lim_{x \to 0^-}e^{\frac{1}{x}} = e^{-\infty} = 0.$
However all of these statements are based on intuition and are not formal. 
I tried to use the epsilon-delta proof in order to show a limit does or doesn't exist.
For the first function I tried using a contradiction in order to show a limit does not exist. So suppose a limit $L$ does exist and $L > 0$. If $L > 0$ than there exists a $\delta > 0$, so that 
$0 < |x - x_0| < \delta \Leftrightarrow 0 < |x - 0| < \delta \Leftrightarrow 0 < x < \delta$ with $|f(x) - L| < \epsilon$. But I don't know how to continue from here. Because normally in an epsilon-delta proof I would now try to find a $\delta$ starting with the term  $|f(x) - L| < \epsilon$ and by using equivalent transformation getting to the term $|x - 0| < \delta$. However I don't know if this is even the right approach in this case. 
For the second function I get stuck at the same point. Is it suggestive to even use the epsilon-delta proof for these kind of functions or is there some completely different solution path that is more suitable.
 A: So I am going to try to give an answer to $(2)$. As already mentioned above, a limit of $\lim_{x \to 0^-}e^{\frac{1}{x}}$ does exist. 
Suppose the limit of $(2)$ is equal to $0$, so $\lim_{x \to 0^-}e^{\frac{1}{x}} = 0.$ I am going to prove this claim by doing an $\epsilon-\delta$ proof.
First of all we need to find a suitable $\delta$. It follows:
$|f(x) - x_0| < \epsilon \\ \Leftrightarrow e^\frac{1}{-x} - 0 < \epsilon \\ \Leftrightarrow e^{-\frac{1}{x}} < \epsilon \\ \Leftrightarrow -\frac{1}{x} < \ln(\epsilon) \\ \Leftrightarrow -1 < \ln(\epsilon)\cdot x \ \ \ (1)\\ \Leftrightarrow -\frac{1}{\ln(\epsilon)} < x \\ \Rightarrow \delta = \frac{1}{\ln(\epsilon)}$
As we found our $\delta$ we can now start with the actual proof.
Let $\epsilon > 0$ be given. Choose $\delta := \frac{1}{\ln(\epsilon)}.$ We need to show from  $|x - x_0| < \delta \Leftrightarrow -x - x_0 < \delta \Leftrightarrow -x - 0 < \delta \Leftrightarrow x > -\delta$ it follows $|f(x) - x_0| < \epsilon.$
$|f(x) - x_0| < \epsilon \\ \Rightarrow e^{-\frac{1}{x}} - 0 = e^{\frac{-1}{x}} \overset{\mathrm{(1)}}{<} e^{\frac{\ln(\epsilon) \cdot x}{x}} = e^{\ln(\epsilon)} = \epsilon \ \ \ \square$
Is the above proof correct? 
