Math Problem Statement

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

Solution

We are tasked with finding the limits of the function f(x)=xe2xf(x) = x \cdot e^{-2x} as xx approaches both negative infinity (xx \to -\infty) and positive infinity (x+x \to +\infty).

1. Limit as x+x \to +\infty

First, let's consider the limit of f(x)=xe2xf(x) = x \cdot e^{-2x} as x+x \to +\infty.

As x+x \to +\infty:

  • e2x0e^{-2x} \to 0 because the exponential function decays very rapidly for large positive xx.
  • x+x \to +\infty.

So, we have an indeterminate form 0\infty \cdot 0. To resolve this, let's rewrite the expression by recognizing that e2xe^{-2x} can be written as 1e2x\frac{1}{e^{2x}}:

f(x)=xe2x=xe2x.f(x) = x \cdot e^{-2x} = \frac{x}{e^{2x}}.

Now, we evaluate:

limx+xe2x.\lim_{x \to +\infty} \frac{x}{e^{2x}}.

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

limx+xe2x=0.\lim_{x \to +\infty} \frac{x}{e^{2x}} = 0.

2. Limit as xx \to -\infty

Now, let's consider the limit of f(x)=xe2xf(x) = x \cdot e^{-2x} as xx \to -\infty.

As xx \to -\infty:

  • e2x=e2xe^{-2x} = e^{2|x|} \to \infty because the exponential grows very rapidly for large negative xx.
  • xx \to -\infty.

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

limxxe2x=.\lim_{x \to -\infty} x \cdot e^{-2x} = -\infty.

Summary of Limits:

limx+xe2x=0,\lim_{x \to +\infty} x \cdot e^{-2x} = 0, limxxe2x=.\lim_{x \to -\infty} x \cdot e^{-2x} = -\infty.

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


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 0\infty \cdot 0?
  3. Can L'Hopital's Rule be applied to this limit problem? If so, how?
  4. What happens to f(x)=xekxf(x) = x \cdot e^{-kx} when kk is a positive constant other than 2?
  5. How does the behavior of f(x)=xekxf(x) = x \cdot e^{kx} change if kk is negative?

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

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits at Infinity
Exponential Functions

Formulas

lim (x → +∞) [x * e^(-2x)] = 0
lim (x → -∞) [x * e^(-2x)] = -∞

Theorems

Indeterminate Forms
Polynomial vs Exponential Growth
L'Hôpital's Rule (Optional for solving indeterminate forms)

Suitable Grade Level

College Level (Calculus I/II)