Evaluation of the indeterminate forms $0^0$, $1^{\infty }$, $\infty ^0$

Given a function of the form

$$f(x)^{\phi(x)}.$$

In order that the function shall take on one of the above three forms, we must have for a certain value of $x$, $f(x) = 0$, $\phi(x) = 0$, giving $0^0$; or, $f(x) = 1$, $\phi(x) = \infty$, giving $1^{\infty }$; or, $f(x) = \infty$, $\phi(x) = 0$, giving ${\infty}^0$. Let $y = f(x)^{\phi(x)}$; taking the logarithm of both sides, $\log y = \phi (x)\log f(x)$. In any of the above cases the logarithm of $y$ (the function) will take on the indeterminate form $0 \cdot \infty$.

Evaluating this by the process illustrated in §13.11 gives the limit of the logarithm of the function. This being equal to the logarithm of the limit of the function, the limit of the function is known.

Example 13.13.1   Evaluate $x^x$ when $x = 0$.

Solution. This function assumes the indeterminate form $0^0$ for $x = 0$. Let $y = x^x$; then $\log y = x \log x = 0 \cdot (-\infty)$, when $x = 0$. By §13.11,

$$\log y \frac{\log x}{\frac{1}{x}} = \frac{-\infty}{\infty},$$

when $x = 0$. By §13.10,

$$\log y \frac{\frac{1}{x}}{-\frac{1}{x^2}} = -x = 0,$$

when $x = 0$. Since $y = x^x$, this gives $\log_e(x^x) = 0$; i.e., $x^x = 1$. Ans.

Example 13.13.2   Evaluate $(1 + x)^{\frac{1}{x}}$ when $x = 0$.

Solution. This function assumes the indeterminate form $1^{\infty }$ for $x = 0$. Let $y = (1 + x)^{\frac{1}{x}}$; then $\log y = \frac{1}{x} \log(1 + x) = \infty \cdot 0$ when $x = 0$. By §13.11, $y = \frac{\log(1 + x)}{x} = \frac{0}{0}$, when $x = 0$. By §13.9, $y = \frac{\frac{1}{1 + x}}{1} = \frac{1}{1 + x} = 1$ when $x = 0$. Since $y = (1 + x)^{\frac{1}{1 + x}}$, this gives $\log_e (1 + x)^{\frac{1}{x}} = 1$; i.e. $(1 + x)^{\frac{1}{x}} = e$. Ans.

Example 13.13.3   Evaluate $\cot x\sin x$ for $x = 0$.

Solution. This function assumes the indeterminate form $\infty ^0$ for $x = 0$. Let $y =\ (\cot x)^{\sin x}$; then $\log y =\ \sin x \log \cot x = 0 \cdot \infty$ when $x = 0$. By §13.11, $\log y = \frac{\log \cot x}{\csc x} = \frac{\infty}{\infty}$ when $x = 0$. §13.10, $\log y = \frac{\frac{-\csc^2 x}{\cot x}}{-\csc x \cot x} = \frac{\sin x}{\cos^2 x} = 0$, when $x = 0$.