# Find $\lim_{x\to-\infty}{x+e^{-x}}$

I have this exercise in my worksheet: $$\lim_{x\to-\infty}{x+e^{-x}}$$ I am always ending up with $-∞+∞$ or $\frac{∞}{∞}$. It says the answer is $+∞$, but how can I get that?

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Do you just want help with the problem in the title or more than one problem? –  DanZimm Jun 5 '13 at 20:03

Negative numbers make me nervous, so let $t=-x$. We want $$\lim_{t\to \infty} (e^t-t).$$ The answer is obvious, $e^t$ is much larger than $t$ if $t$ is large. If you want to be formal, after a (short) while $t\lt \frac{e^t}{2}$, so after a short while $e^t-t\gt \frac{1}{2}e^t$.

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Using $e^x\ge 1+x$ for all $x\in\mathbb R$, we find for $x\ge-1$ that $e^x=(e^{x/2})^2\ge(1+\frac x2)^2= 1+x+\frac14x^2$, hence $$\lim_{x\to-\infty}(x+e^{-x})=\lim_{x\to+\infty}(-x+e^{x}) \ge\lim_{x\to+\infty}(1+\frac14x^2)=+\infty.$$

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Here is a textbook way to solve this problem using L'Hopital's Rule:

$$\lim_{x\to-\infty}x+e^{-x}=\lim_{x\to-\infty}\frac{xe^x+1}{e^x}$$

For the numerator we have:

$$\lim_{x\to-\infty}xe^x=\lim_{x\to-\infty}\frac{x}{e^{-x}}=\lim_{x\to-\infty}\frac{1}{-e^{-x}}=0$$

It follows that the numerator approaches $1$, and the denominator approaches $0$ from the right. The limit is thus $+\infty$.

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may you give an upvote in order for me to vote for your answer since I have < 15 reputation so upvote in not allowed :D –  A.Jouni Jun 5 '13 at 20:25

"$-\infty+\infty$" is an indeterminate form, meaning the limit could be any finite number or could be $+\infty$ or could be $-\infty$, depending on what functions you're working with.

Look at what happens when $x$ goes from $-100$ to $-101$, one step closer to $-\infty$. Then $e^{-x}$ gets multiplied by $e$, so it gets to be more than two-and-a-half times as big. But the other term, $x$ decreases by only one. The result is that the function gets immensely bigger, in the "$+$"-direction.

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$x+e^{-x}=1+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots$ which clearly tends to infinity as $x\to -\infty$.

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I'll assume $x\in \mathbb{R}$. Then the continuous function $f(x) = x+e^{-x}$ goes to $+\infty$ as $x\rightarrow -\infty$. This can visually be seen by graphing the function, but here is a rigorous proof:

Since $f$ is continuous, then $lim_{x\rightarrow y} f(x) = f(y)$, and since $e^x$ increases faster than any power of $x$ (see baby Rudin (chapter 8) for this fun fact), it follows that $\lim_{x\rightarrow - \infty} f(x) = +\infty$. This fact cited from Rudin justifies why $e^z$ dominates $z$ in the expression $e^z -z$ and thus why $e^z - z$ goes to $+\infty$ as $z\rightarrow +\infty$, i.e. why $f(x)\rightarrow +\infty$ as $x\rightarrow -\infty$.

Corollary from the facts presented herein: let $n>0$ be a positive integer. Then $$x^n + e^{-x}$$ goes to $+\infty$ as $x\rightarrow -\infty$. There are obviously several other generalizations.

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This is wrong. $e^x$ isn't the fastest increasing continuous function...there is no such function. $e^{e^x}$ for instance increases faster. –  JLA Jun 5 '13 at 22:43
you're right. i've edited –  doncorleone Jun 5 '13 at 22:56
don corleone expresses his sincerest apologies and now claims to have one of the best answers here –  doncorleone Jun 5 '13 at 22:58

The given expression does not have a limit. That is to say, it does not converge on a value as $x$ grows large, negative.

As $x$ grows large and negative, the $x$ term grows large and negative. The $e^{-x}$ term grows large and positive, and its growth is faster than that of the $x$ term, because it is an exponential function whereas $x$ is polynomial. Exponentials grow faster than polynomials of all orders. Therefore, the exponential term dominates and the overall expression grows large, and positive.

However, it does so without limit. We cannot say that the limit is $+\infty$, because that is nonsense. A limit, when it exists, is a number, whereas $+\infty$ is not a number and does not substitute for a limit that doesn't exist.

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Even though the limit doesn't exist, it's still the case that improper limits are both common and useful. Saying that $f(x) \to \infty$ as $x\to\infty$ gives a lot more information about the behaviour of $f$ than just saying that the limit doesn't exist. –  mrf Jun 5 '13 at 21:05
@mrf but $f(x)\rightarrow\infty$ as $x\rightarrow\infty$ is not exactly the same as $\displaystyle\lim_{x\to\infty}f(x) = \infty$. –  Kaz Jun 5 '13 at 21:57
@Kaz what you are saying goes against convention, it is common to write things like $\lim\limits_{x\to\infty}f(x)=\infty$. –  JLA Jun 5 '13 at 22:46
Well, it is wrong, because you're writing an equation whose right hand side is infinity. Convention be damned. There is also a convention that dividing by inifinity neatly yields zero, and vice versa. If you want to write that, you have to clarify that you're giving a different meaning to the $=$ symbol which is tied to the limit notation on the far left, and not forming an equation. –  Kaz Jun 5 '13 at 22:48
@Kaz: I don't think that needs a special convention. The common convention says that $$\lim_{x\to\infty}x^2=\infty$$ is valid (and understood). Limits are usually understood to be taken in the two-point compactification of $\mathbb{R}$ or the one-point compactification of $\mathbb{C}$. –  robjohn Jun 5 '13 at 23:32