- Thread starter
- #1
Petrus
Well-known member
- Feb 21, 2013
- 739
Hello MHB,
\(\displaystyle \lim_{x->\pm\infty}xe^{\frac{2}{x}}-x\)
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,
\(\displaystyle |\pi\rangle\)
\(\displaystyle \lim_{x->\pm\infty}xe^{\frac{2}{x}}-x\)
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,
\(\displaystyle |\pi\rangle\)