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?
 A: 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
$$ \lim_{x\to-\infty}(x+e^{-x})=\lim_{x\to+\infty}(-x+e^{x})  \ge\lim_{x\to+\infty}(1+\frac14x^2)=+\infty.$$
A: 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 $+\infty$.
A: "$-\infty+\infty$" 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.
A: $x+e^{-x}=1+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots$ which clearly tends to infinity as $x\to -\infty$.
A: 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 $e^t-t\gt \frac{1}{2}e^t$.
A: 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.
A: 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. 
