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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

\(\displaystyle \lim_{x->\pm\infty}xe^{\frac{2}{x}}-x\)

I start to divide by x and we know that \(\displaystyle \lim_{x->\pm\infty} \frac{2}{x}=0\)

with other words we get \(\displaystyle 1-1=0\) but that is wrong, how do I do this

Regards,

\(\displaystyle |\pi\rangle\)