Calculating the limit of $(x-1)e^{1/x}-x$ I've had some trouble calculating the following lim (which equals zero):
$$\lim\limits_{x \to \infty} {(x-1)}{e}^{1/x}-x$$
I always end up at $\infty-\infty$ which is undefined
Help much appreciated!
 A: Write it as 
$$x(e^{\frac{1}{x}}-1)-e^{\frac{1}{x}}$$
Now the individulal terms both go to $1$. The first being an application of $$\lim\limits_{y\to 0}\frac{e^y-1}{y}=1$$
A: With 
$$ e^{1/x} = 1 + \frac1x +O\Big(\frac1{x^2}\Big) $$
you will find $0$ as the limit.
A: $1)$ $t=1/x$  
$2)$ L'Hôpital's rule
$\lim\limits_{x \to \infty} {(x-1)}{e}^{1/x}-x = \lim\limits_{t \to 0} {(1/t-1)}{e}^{t}-1/t= \lim\limits_{t \to 0} \dfrac{(1-t)e^t-1}{t}=\lim\limits_{t \to 0} \dfrac{-e^t+(1-t)e^t}{1}=0$
A: Let $x=\frac1y$ with $y\to 0^+$
$${(x-1)}{e}^{1/x}-x=\frac{1-y}{y}e^y-\frac1y=\frac{e^y-1}{y}-e^y\to1-1=0$$
A: $(x-1)e^{1/x} - x+1-1=$
$(x-1)(e^{1/x}-1) -1=$
$x(1-1/x)(e^{1/x}-1) -1=$
$(1-1/x)\dfrac{e^{1/x}-1}{1/x}-1;$
Set $y=1/x$ , and consider $y \rightarrow 0^+$.
$\lim_{ y \rightarrow 0^+} (1-y)\dfrac{e^y-1}{y} -1=$
$\lim_{y \rightarrow 0^+}(1-y)×$
$\lim_{y \rightarrow 0^+}\dfrac{e^y-1}{y} -1=$
$1×1-1=0.$
Used :
$\lim_{y \rightarrow 0^+}\dfrac{e^y-1}{y}= (e^y)'_{y=0}=1.$
A: Also, the L'Hospital works. 
It's
$$\lim_{x\rightarrow0^+}\frac{(1-x)e^x-1}{x}=\lim_{x\rightarrow0^+}(-e^x+1-x)=0.$$
A: Since $1+z < e^x < \dfrac1{1-z}$
for $0<z<1$,
$1+1/x <e^{1/x}<\dfrac1{1-1/x}
=(x-1)/x$
so the expression is between
$-1/x$ and $0$ for $x>1$.
