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

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    Infinity Limit
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Discussion Overview

The discussion revolves around evaluating the limit $$\lim_{x->\pm\infty}xe^{\frac{2}{x}}-x$$ using L'Hopital's Rule and other mathematical techniques. Participants explore different approaches to simplify and solve the limit, addressing potential pitfalls in the reasoning.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • One participant initially attempts to evaluate the limit by dividing by x and concludes that it approaches 0, expressing confusion about this result.
  • Another participant suggests rewriting the limit as $$\dfrac{e^{2/x}-1}{1/x}$$ to facilitate evaluation.
  • The original poster later claims to have solved the limit, arriving at an answer of 2, indicating a change in understanding.
  • A further contribution expands on the limit evaluation by expressing it in terms of a series expansion, showing the steps leading to the limit's behavior as x approaches infinity.

Areas of Agreement / Disagreement

There is no clear consensus on the initial approach to the limit, as participants present different methods and interpretations. The discussion includes both a claimed solution and an alternative expansion approach, indicating multiple perspectives on the problem.

Contextual Notes

Some assumptions regarding the behavior of exponential functions at infinity and the application of L'Hopital's Rule are present but not fully resolved. The discussion reflects varying levels of understanding and interpretation of the limit's evaluation.

Petrus
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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 :confused:

Regards,
$$|\pi\rangle$$
 
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Re: limit

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

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

Regards,
$$|\pi\rangle$$
 
Re: limit

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

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

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