Calculating the limit of $(x-1)e^{1/x}-x$

Question

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!

Answer 1

$1)$ $t=1/x$

$2)$ L'Hôpital's rule

$\lim\limits_{x \to \infty} {(x-1)}{e}^{1/x}-x = \lim\limits_{t \to 0} {(1/t-1)}{e}^{t}-1/t= \lim\limits_{t \to 0} \dfrac{(1-t)e^t-1}{t}=\lim\limits_{t \to 0} \dfrac{-e^t+(1-t)e^t}{1}=0$

Answer 2

With $$ e^{1/x} = 1 + \frac1x +O\Big(\frac1{x^2}\Big) $$ you will find $0$ as the limit.

Answer 3

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 $$\lim\limits_{y\to 0}\frac{e^y-1}{y}=1$$

Answer 4

Also, the L'Hospital works.

It's $$\lim_{x\rightarrow0^+}\frac{(1-x)e^x-1}{x}=\lim_{x\rightarrow0^+}(-e^x+1-x)=0.$$

Answer 5

Let $x=\frac1y$ with $y\to 0^+$

$${(x-1)}{e}^{1/x}-x=\frac{1-y}{y}e^y-\frac1y=\frac{e^y-1}{y}-e^y\to1-1=0$$

Answer 6

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 $x>1$.

Answer 7

$(x-1)e^{1/x} - x+1-1=$

$(x-1)(e^{1/x}-1) -1=$

$x(1-1/x)(e^{1/x}-1) -1=$

$(1-1/x)\dfrac{e^{1/x}-1}{1/x}-1;$

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 :

$\lim_{y \rightarrow 0^+}\dfrac{e^y-1}{y}= (e^y)'_{y=0}=1.$