Problem 43 - Precalculus

Question

For the following exercises, use a graphing utility to find numerical or graphical evidence to determine the left and righthand limits of the function given as \(x\) approaches \(a\). If the function has a limit as \(x\) approaches \(a\), state it. If not, discuss why there is no limit. $$\lim _{x \rightarrow 0} e^{-\frac{1}{x^{2}}}$$

Step-by-Step Solution

Answer

The limit is 0 as both the left and right-hand limits approach zero.

Step 1: Understanding the Function

The function given is \( f(x) = e^{-\frac{1}{x^2}} \). As \( x \) approaches 0, we analyze the expression \( \frac{1}{x^2} \). This expression becomes very large for values of \( x \) close to zero but not equal to zero.

Step 2: Left-Hand Limit Analysis

For \( x \to 0^- \), where \( x \) approaches zero from the left (negative direction), \( \frac{1}{x^2} \) still tends to infinity because it is squared, making it positive regardless of the sign of \( x \). Consequently, \( e^{-\frac{1}{x^2}} = e^{large\ negative\ number} \) approaches zero.

Step 3: Right-Hand Limit Analysis

Similarly, for \( x \to 0^+ \), where \( x \) approaches zero from the right (positive direction), \( \frac{1}{x^2} \) again tends to infinity, resulting in \( e^{-\frac{1}{x^2}} = e^{large\ negative\ number} \) which also approaches zero.

Step 4: Conclusion on the Limit

Since both the left-hand limit and the right-hand limit as \( x \rightarrow 0 \) approach zero, the overall limit of the function \( \lim_{x \to 0} e^{-\frac{1}{x^2}} \) exists and is zero.

Key Concepts

Left-Hand Limit

When we talk about the left-hand limit of a function, we're understanding how the function behaves as it approaches a particular value from the left side. In mathematical terms, this is sometimes expressed as \( x \to a^- \). For the function \( f(x) = e^{-\frac{1}{x^2}} \), let's see what happens when \( x \to 0^- \), meaning from negative values near zero.

Firstly, notice that for any negative \( x \), \( \frac{1}{x^2} \) becomes positive because squaring removes any negative sign. As \( x \) gets closer and closer to zero, \( \frac{1}{x^2} \) grows very large because you're effectively dividing 1 by an extremely small number. In exponential terms, when you have \( e^{-\text{large number}} \), it approaches zero. Therefore, the left-hand limit of our function as \( x \to 0^- \) is 0.

• Approaching from the negative side means we consider values smaller than the target (zero).
• Squaring a negative number results in a positive.
• The larger the exponent's negative value, the closer \( e \) gets to zero.

Right-Hand Limit

The right-hand limit refers to what happens as the function closes in on a specific point from the right side or positive direction. This is written as \( x \to a^+ \). In examining the function \( f(x) = e^{-\frac{1}{x^2}} \), we want to understand its behavior as \( x \) approaches zero from positive values.

As before, with \( x \) being positive and closing in on zero, \( \frac{1}{x^2} \) again becomes a very large number. But since \( x \) is positive, \( \frac{1}{x^2} \) stays positive. So, \( e^{-\text{large positive number}} \) still approaches zero. This mirrors the behavior observed with the left-hand limit, which signifies that coming from the right, the function also tends towards zero.

• Consider values greater than zero for the approach from the right.
• Positive numbers squared remain positive, increasing \( \frac{1}{x^2} \).
• As the exponent becomes a large negative, the function value diminishes to zero.

Exponential Functions

Exponential functions are a crucial part of mathematics, especially in understanding growth and decay processes. They have the form \( f(x) = a^{x} \), where \( a \) is a positive constant. With our specific function, \( f(x) = e^{-\frac{1}{x^2}} \), it suggests an exponential decay.

The base \( e \) is a special constant (\( \approx 2.71828 \)) naturally arising in growth rates and compound interest. But, in our example, the exponent is particularly notable. We are raising \( e \) to a power that's negative and dependent on \( x \). Given that \( -\frac{1}{x^2} \) accelerates towards a large negative value as \( x \) nears zero, the function \( e^{\text{negative large number}} \) rapidly declines towards zero.

• Exponential functions can model rapid growth when the exponent is positive.
• For negative exponents, they model rapid decay, approaching zero.
• The nature of this decline in our function illustrates foundational exponential decay behavior.