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I thought the answer to this is just 0. However, when I plug this into online calculators, they say it is not possible to evaluate or it is math error.

$$\lim_{x\to\infty}2e^x-\frac{2}{x}-x^2$$

I simply do not understand. The third term = infinity. The second term = 0. The 1st term = infinity. Hence the overall result should just be 0. Why do I get math error??

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    $\begingroup$ Try plotting this function, and see if you can guess what it's limit is. $\endgroup$ Commented Aug 28, 2023 at 2:09
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    $\begingroup$ This limit does not exist. $\lim_{x\to-\infty}2e ^{x}-\frac{2}{x}-x^2=0-0-\infty=-\infty$; $\lim_{x\to+\infty}2e ^{x}-\frac{2}{x}-x^2=+\infty$. By the way, there's no way saying "infinity-infinity = 0". $\infty-\infty$ is simply undefined, you have to consider the particualr expression to determine its value. $\endgroup$
    – Ricky
    Commented Aug 28, 2023 at 2:17
  • $\begingroup$ If you are familiar with derivatives: Let $~f(x) = 2e^x - x^2 \implies f'(x) = 2e^x - 2x.~$ Clearly, as $~x \to \infty, ~f'(x) \to \infty,~$ so $~f(x)~$ must also go to $~\infty.$ $\endgroup$ Commented Aug 28, 2023 at 3:26
  • $\begingroup$ Thanks Ricky, that makes sense. $\endgroup$ Commented Aug 30, 2023 at 1:05

2 Answers 2

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As you said, the second term is 0. Your mistake lies in that you said, since the first term and third terms are infinity, they cancel out and give 0. This is in general not true. For example, consider the limit $$\lim_{x \to \infty} 2x - x.$$ This limit is not 0, even though the limit of each individual term is infinity.

The point here is that you need to think about whether the $2e^x$ term or the $x^2$ grows faster.

Here is one way you can think about it. What is the limit of $\frac{2e^x}{x^2}$? If this limit is infinity, this is one way you can conclude that $2e^x$ grows faster than $x^2.$ Otherwise if this limit is 0, then it is the other way around.

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    $\begingroup$ To be clear, even the fact that the ratio tends to 1 is also insufficient- $\frac{x+\ln(x)}{x}$ tends to $1$, but $\lim_{x\to\infty} (x+\ln(x))-(x)\to\infty$ nevertheless. $\endgroup$ Commented Aug 28, 2023 at 2:57
  • $\begingroup$ Good point, thanks for that note! $\endgroup$
    – Alan Chung
    Commented Aug 28, 2023 at 3:03
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If we open the exponential: \begin{align*} \lim_{x\to \infty} 2e^x-\frac{2}{x}-x^2&=\lim_{x\to \infty} 2e^x-x^2\\ &=\lim_{x\to \infty} 2(1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...)-x^2\\ &=\lim_{x\to \infty} 2(1+x+\frac{x^3}{3!}+...)\\ &\to \infty. \end{align*}

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