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Limit Involving Infinity

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 :confused:

Regards,
\(\displaystyle |\pi\rangle\)
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
Re: limit

Rewrite: \(\displaystyle \dfrac{e^{2/x}-1}{1/x}\)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: limit

Rewrite: \(\displaystyle \dfrac{e^{2/x}-1}{1/x}\)
Thanks solved it now!:) got 2 as answer now!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: limit

\(\displaystyle x(e^{\frac{2}{x}}-1) = x ( 1+\frac{2}{x}+\frac{2}{x^2}+\mathcal{o}(\frac{1}{x^2})-1 )\)

\(\displaystyle x( \frac{2}{x}+\frac{2}{x^2}+\mathcal{o} (\frac{1}{x^2}) ) = 2+\frac{2}{x}+\mathcal{o} (\frac{1}{x}) \)