MHB How to Evaluate a Limit Involving Infinity using L'Hopital's Rule

[Petrus]

Hello MHB,
$$\lim_{x->\pm\infty}xe^{\frac{2}{x}}-x$$
I start to divide by x and we know that $$\lim_{x->\pm\infty} \frac{2}{x}=0$$
with other words we get $$1-1=0$$ but that is wrong, how do I do this

Regards,
$$|\pi\rangle$$

 

[tkhunny]

Re: limit

Rewrite: $$\dfrac{e^{2/x}-1}{1/x}$$

 

[Petrus]

Re: limit

Rewrite: $$\dfrac{e^{2/x}-1}{1/x}$$

Thanks solved it now!:) got 2 as answer now!:)

Regards,
$$|\pi\rangle$$

 

[alyafey22]

Re: limit

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

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