Zero raised to the power zero form
Introduction to the Indeterminate Form \(0^0\)
Consider the limit
where \( f(x) \to 0^+ \) (not \( 0^- \), since the base of an exponent must be positive) and \( g(x) \to 0 \). This results in the indeterminate form \( 0^0 \).
To see why this is indeterminate, we rewrite the expression using exponentials:
Now, since \( f(x) \to 0^+ \), we have \( \ln f(x) \to -\infty \). Thus, the exponent
takes the form \( 0 \times \infty \), which is an indeterminate form. This confirms that the original limit cannot be determined directly and requires further analysis.
Intuitively, the indeterminacy of \( 0^0 \) arises from two conflicting tendencies:
- The base \( f(x) \to 0^+ \), which tends to pull the expression towards \( 0 \).
- The exponent \( g(x) \to 0 \), which tends to pull the expression towards \( 1 \), since any number raised to the power of \( 0 \) is \( 1 \).
Because these opposing forces act simultaneously, the final value depends on the relative rates at which \( f(x) \to 0 \) and \( g(x) \to 0 \). Different functions can lead to different outcomes, making \( 0^0 \) an indeterminate form.
Evaluation
To evaluate limits of the form
where \( f(x) \to 0^+ \) and \( g(x) \to 0 \), resulting in the indeterminate form \( 0^0 \), we rewrite the expression using exponentials:
Now, the exponent inside the limit is
Since \( f(x) \to 0^+ \), we have \( \ln f(x) \to -\infty \), while \( g(x) \to 0 \), making the exponent take the indeterminate form \( 0 \times (-\infty) \), which is equivalent to \( 0 \times \infty \).
To resolve this, we transform \( g(x) \ln f(x) \) into \( \frac{\infty}{\infty} \) form by rewriting as:
This can then be evaluated using L'Hôpital's Rule, algebraic techniques etc. Once the limit of the exponent is determined, we exponentiate the result to obtain
Thus, the key idea is to convert the \( 0^0 \) form into an exponential expression where the exponent follows the \( 0 \times \infty \) pattern, which can be handled using known techniques.