如何仅用简单代数运算或常用极限列表求解limₓ→-∞ xeˣ=0

Hey there! Let's work through this limit step by step without relying on fancy tools like L'Hospital's Rule or Taylor series—just good old algebra and some standard limit facts we already know.

Step 1: Variable Substitution to Flip the Infinity Direction #

First, let's make a substitution to turn the negative infinity limit into a positive one, which is easier to reason about. Let $t = -x$. When $x \to -\infty$, $t$ will approach $+\infty$. Plugging this into the original limit:
$$
\lim_{x\to -\infty} xe^x = \lim_{t\to +\infty} (-t)e^{-t} = -\lim_{t\to +\infty} \frac{t}{e^t}
$$
Now we just need to show that $\lim_{t\to +\infty} \frac{t}{e^t} = 0$, and the original limit will follow directly.

Step 2: Use a Fundamental Exponential Inequality #

A key basic fact about the exponential function is that for any $x > 0$, $e^x > 1 + x$. This comes from the definition of $e^x = \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n$—when you expand the right-hand side with the binomial theorem, all extra terms are positive, so the limit is strictly greater than $1 + x$.

For $t > 0$, let's split $e^t$ into a product of two identical exponentials: $e^t = e^{t/2} \cdot e^{t/2}$. Applying the inequality above to each $e^{t/2}$ (since $t/2 > 0$ when $t>0$):
$$
e^t > \left(1 + \frac{t}{2}\right)^2
$$
Expanding the right-hand side gives:
$$
\left(1 + \frac{t}{2}\right)^2 = 1 + t + \frac{t^2}{4}
$$
For $t > 2$, the quadratic term $\frac{t^2}{4}$ is positive and dominates the lower-degree terms, so we can simplify the inequality to:
$$
e^t > \frac{t^2}{4}
$$
Taking reciprocals (and reversing the inequality, since both sides are positive):
$$
\frac{1}{e^t} < \frac{4}{t^2}
$$
Multiply both sides by $t$ (which is positive, so the inequality direction stays the same):
$$
\frac{t}{e^t} < \frac{4}{t}
$$

Step 3: Apply the Squeeze Theorem #

We know that for all $t > 0$, $\frac{t}{e^t}$ is positive (since both $t$ and $e^t$ are positive). So we have the following bounds:
$$
0 < \frac{t}{e^t} < \frac{4}{t}
$$
As $t \to +\infty$, $\frac{4}{t}$ approaches 0. By the Squeeze Theorem (also called the Sandwich Theorem), $\frac{t}{e^t}$ must also approach 0. That means:
$$
\lim_{t\to +\infty} \frac{t}{e^t} = 0
$$

Step 4: Wrap Up the Original Limit #

Going back to our substituted expression from Step 1:
$$
\lim_{x\to -\infty} xe^x = -\lim_{t\to +\infty} \frac{t}{e^t} = -0 = 0
$$

That's all there is to it! Every step uses basic substitution, a fundamental exponential inequality, and the Squeeze Theorem—no advanced calculus tricks required.

内容的提问来源于stack exchange，提问作者math_mx