Limit Involving Infinity

[Petrus]

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

 

[tkhunny]

Re: limit

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

 

[Petrus]

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]

Re: limit

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

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