Evaluate $\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$ I want to find the following limit:
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$$
This is what I do. I change the variable $t=-x$ and I have the following limit:
$$\lim_{t\rightarrow +\infty}{e^{\frac {1}{2+t}}\cdot{\frac{t^2-2t-1}{-t-2}}+t}=\lim_{t\rightarrow+\infty}{h(x)}$$
We have $e^{\frac{1}{2+t}}\rightarrow1$ for $t\rightarrow+\infty$
Therefore I think (this is the passage I'm less sure about)
$$h(x)\sim \frac{t^2-2t-1}{-t-2}+t=\frac{t^2-2t-1+t(-t-2)}{-t-2}=\frac{t^2-2t-1-t^2-2t}{-t-2}=\frac{-4t-1}{-t-2}\sim{\frac {-4t}{-t}}\rightarrow4$$
The solution should actually be $3$. Any hints on what I'm doing wrong?
 A: Let set $\frac {1}{2-x}=y \to 0$
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x=\lim_{y\rightarrow 0}{e^{y}\cdot\frac{\frac1{y^2}-\frac 6 y+7}{-\frac1y}}-2+\frac1y$$$$=\lim_{y\rightarrow 0}e^{y}\cdot\left(-\frac1{y}+6-7y\right)-2+\frac1y$$
then use $e^y=1+y+o(y)$ to obtain
$$(1+y+o(y))\cdot\left(-\frac1{y}+6-7y\right)-2+\frac1y$$
$$-\frac1{y}+6-7y-1+6y-7y^2+o(1)-2+\frac1y=3+o(1) \to 3$$
A: I think the first substitution isn't too useful, but let's keep it nontheless. $$e^{1/(2+x)}\frac{-x^2+2x+1}{x+2}+x=(4-x)e^{1/(2+x)}-7\frac{e^{1/(2+x)}}{x+2}+x=\\=4e^{1/(x+2)}-\frac7{x+2}e^{1/(2+x)}+xe^{1/(2+x)}\left(e^{-1/(2+x)}-1\right)=\\=4e^{1/(x+2)}-\frac7{x+2}e^{1/(2+x)}-\frac x{x+2}e^{1/(2+x)}\frac{e^{-1/(2+x)}-1}{-\frac1{2+x}}$$
Now, since $-\frac1{x+2}\to 0^-$ as $x+2\to\infty$ and $\lim_{s\to 0}\frac{e^s-1}s=1$, $$\lim_{x\to\infty}4e^{1/(x+2)}-\frac7{x+2}e^{1/(2+x)}-\frac x{x+2}e^{1/(2+x)}\frac{e^{-1/(2+x)}-1}{-\frac1{2+x}}=\\=\left[4e^0-\frac7\infty e^0-1\cdot e^0\cdot 1\right]=3$$
Now that you should know one more idea than before, you can see for yourself how your initial guess fares against $$\lim_{x\to\infty} xe^{1/(\alpha +\beta x)}-x,\qquad \beta\ne 0$$ as opposed to the correct answer.
A: ALternatively:
$$\lim_{x\rightarrow -\infty}\left[{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x\right]=\\
\lim_{x\to-\infty} \left[e^{\frac{1}{2-x}} \left(x+4+\frac7{x-2}\right)-x\right]=\\
\lim_{x\to-\infty} \left[e^{\frac{1}{2-x}}\left(x+4\right)-x\right] +\underbrace{\lim_{x\to-\infty} e^{\frac{1}{2-x}} \cdot \frac7{x-2}}_{=0}=\\
\lim_{x\to-\infty} \left[e^{\frac{1}{2-x}}\left(x+4\right)-(x+4)+4\right]=\\
\lim_{x\to-\infty} \left[\frac{e^{\frac1{2-x}}-1}{\frac1{2-x}}\cdot \frac{x+4}{2-x}+4\right]=\\
1\cdot (-1)+4=3,$$
where it was used:
$$\lim_{x\to 0} \frac{e^x-1}{x}=1.$$
A: Although $\lim\limits_{x\to-\infty}e^{\frac1{2-x}}=1$, we cannot simply replace it by $1$ in the limit because it is actually $1+\frac1{2-x}+O\!\left(\frac1{(2-x)^2}\right)$

For $x$ near $\pm\infty$,
$$
\begin{align}
e^{\frac1{2-x}}\frac{x^2+2x-1}{x-2}-x
&=\overbrace{\left(1+\frac1{2-x}+O\!\left(\frac1{(2-x)^2}\right)\right)}^{e^{\frac1{2-x}}}\overbrace{\left(x+4-\frac7{2-x}\right)\vphantom{\left(\frac1{()^2}\right)}}^{\frac{x^2+2x-1}{x-2}}-x\\
&=\frac x{2-x}+4+O\!\left(\frac1{2-x}\right)
\end{align}
$$
Now let $x\to-\infty$.

Simply replacing $e^{\frac1{2-x}}$ by $1$ misses the $\frac x{2-x}$ in the formula above and leaves us only with $4$ in the limit.
