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?

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.
