Find $\lim_{x\to-\infty}{x+e^{-x}}$
Solution 1:
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 $e^t-t\gt \frac{1}{2}e^t$.
Solution 2:
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 $$ \lim_{x\to-\infty}(x+e^{-x})=\lim_{x\to+\infty}(-x+e^{x}) \ge\lim_{x\to+\infty}(1+\frac14x^2)=+\infty.$$