solve limit when x goes minus infinite and pus infinite and function is x*e^(-2x)

We are tasked with finding the limits of the function \( f(x) = x \cdot e^{-2x} \) as \( x \) approaches both negative infinity (\( x \to -\infty \)) and positive infinity (\( x \to +\infty \)).

### 1. Limit as \( x \to +\infty \)

First, let's consider the limit of \( f(x) = x \cdot e^{-2x} \) as \( x \to +\infty \).

As \( x \to +\infty \):
- \( e^{-2x} \to 0 \) because the exponential function decays very rapidly for large positive \( x \).
- \( x \to +\infty \).

So, we have an indeterminate form \( \infty \cdot 0 \). To resolve this, let's rewrite the expression by recognizing that \( e^{-2x} \) can be written as \( \frac{1}{e^{2x}} \):

\[
f(x) = x \cdot e^{-2x} = \frac{x}{e^{2x}}.
\]

Now, we evaluate:

\[
\lim_{x \to +\infty} \frac{x}{e^{2x}}.
\]

This is a case of the limit \( \frac{\text{polynomial}}{\text{exponential}} \), where the exponential function grows much faster than the polynomial. Therefore, this limit is:

\[
\lim_{x \to +\infty} \frac{x}{e^{2x}} = 0.
\]

### 2. Limit as \( x \to -\infty \)

Now, let's consider the limit of \( f(x) = x \cdot e^{-2x} \) as \( x \to -\infty \).

As \( x \to -\infty \):
- \( e^{-2x} = e^{2|x|} \to \infty \) because the exponential grows very rapidly for large negative \( x \).
- \( x \to -\infty \).

Thus, the function becomes \( (-\infty) \times \infty \), which suggests that the limit will be negative infinity:

\[
\lim_{x \to -\infty} x \cdot e^{-2x} = -\infty.
\]

### Summary of Limits:
\[
\lim_{x \to +\infty} x \cdot e^{-2x} = 0,
\]
\[
\lim_{x \to -\infty} x \cdot e^{-2x} = -\infty.
\]

Let me know if you'd like further details or have any questions!

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Here are 5 follow-up questions:
1. How does the behavior of exponential functions influence the limits of functions involving products?
2. What other methods can we use to solve indeterminate forms like \( \infty \cdot 0 \)?
3. Can L'Hopital's Rule be applied to this limit problem? If so, how?
4. What happens to \( f(x) = x \cdot e^{-kx} \) when \( k \) is a positive constant other than 2?
5. How does the behavior of \( f(x) = x \cdot e^{kx} \) change if \( k \) is negative?

**Tip:** For limits involving both polynomials and exponentials, exponentials generally dominate as \( x \to +\infty \).

Solve limit when x goes to minus infinity and plus infinity for the function x * e^(-2x).

Explore how to solve the limits of the function f(x) = x * e^(-2x) as x approaches both positive and negative infinity. This problem demonstrates key calculus concepts such as indeterminate forms and the comparison of polynomial and exponential growth rates. Suitable for Calculus students at the college level.

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