<s>[INST] 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?
 [/INST] 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:
$$\lim_{x\to 0} \frac{e^x-1}{x}=1.$$
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