find limit without L'hospital if possible I'm struggling with $\displaystyle \lim_{x \to 0} \frac{x-1}{x} \cdot e^{-1/|x|}$ so I want to find 
$\displaystyle \lim_{x \to 0^+} \frac{x-1}{x} \cdot e^{-1/x}$ and 
$\displaystyle \lim_{x \to 0^-} \frac{x-1}{x} \cdot e^{1/x}$
but I have no idea how to attack this
 A: Since $\left(1+\frac{1/|x|}{2}\right)^2\leq e^{1/|x|}$ and $\left|1-x\right|\leq2$ for $|x|<1$ then
$$0\leq\left|\frac{x-1}{x}e^{-1/|x|}\right|\leq2\frac{|x|}{\left(|x|+\frac{1}{2}\right)^2}\stackrel{x\to0}{\longrightarrow}0$$
A: Informally:
$$\lim_{x\to 0+}\frac{x-1}{x}e^{-1/x}=\lim_{x\to 0+}\left(1-\frac{1}{x}\right)e^{-1/x}=\lim_{y\to\infty}\left(1-y\right)e^{-y}=0$$
Since the exponential $e^{-y}$ goes to $0$ much faster than $y$ can grow.
And similarly:
$$\lim_{x\to 0-}\frac{x-1}{x}e^{1/x}=\lim_{x\to 0-}\left(1-\frac{1}{x}\right)e^{1/x}=\lim_{y\to-\infty}\left(1-y\right)e^{y}=0$$
A: No need to consider the two sides. 
$$
\lim_{x \to 0} \left|\frac{x-1}{x} \cdot e^{-1/|x|}\right|=\lim_{x \to 0}|x-1|\cdot\lim_{x \to 0}\frac{1}{|x|e^{1/|x|}}=\lim_{x \to 0}\frac{1}{|x|e^{1/|x|}}
$$
Since $e^{1/|x|}>1/2|x|^2$, 
$$
\lim_{x \to 0}\frac{1}{|x|e^{1/|x|}}\leq \lim_{x \to 0}\frac{1}{|x|/2|x|^2}=\lim_{x \to 0}2|x|=0
$$
Therefore:
$$
\lim_{x \to 0} \left|\frac{x-1}{x} \cdot e^{-1/|x|}\right|=0
$$
and consequently
$$
\lim_{x \to 0}\frac{x-1}{x} \cdot e^{-1/|x|}=0
$$
A: we will assume $$ \lim_{u\to\infty} \frac{u}{e^u} = 0.$$
we can look at $$ \lim_{x \to 0+}\frac{x-1}{x}e^{-1/x} = \lim_{x \to 0+}\frac{-1}{x}e^{-1/x} = \lim_{u \to \infty}-\frac{u}{e^u} = 0.$$ 
we made the substitution $u = 1/x.$
in the same way
$$ \lim_{x \to 0-}\frac{x-1}{x}e^{1/x} = \lim_{x \to 0-}\frac{-1}{x}e^{-1/x} = \lim_{u \to \infty}\frac{u}{e^u} = 0.$$ 
