Evaluation of Infinity raised to power zero form

Introduction to the Indeterminate Form \(\infty^0\)

Consider the limit

\[ \lim_{x \to a} f(x)^{g(x)} \]

where \( f(x) \to \infty \) and \( g(x) \to 0 \). This results in the indeterminate form \(\infty^0\).

To see why this is indeterminate, rewrite the expression using exponentials:

\[ \lim_{x \to a} f(x)^{g(x)} = \lim_{x \to a} e^{g(x) \ln f(x)}. \]

Now, examine the exponent:

\[ \lim_{x \to a} g(x) \ln f(x). \]

Since \( g(x) \to 0 \) and \( \ln f(x) \to \infty \), the exponent takes the form \(0 \times \infty\), which is itself indeterminate. Thus, the original limit is also indeterminate.

Alternatively, we can understand the indeterminacy intuitively:

• The base \( f(x) \) is growing to infinity, which tends to pull the value of the function towards infinity.
• The exponent \( g(x) \) is shrinking to zero, which tends to pull the value towards 1, since any number raised to the power of 0 is 1.

These are two opposing tendencies, and the final result depends on how quickly \( f(x) \) approaches infinity relative to how quickly \( g(x) \) approaches zero. The limit can take any real value, depending on the behavior of \( g(x) \ln f(x) \).

How to evaluate infinity raised to power zero form

To evaluate limits of the form

\[ \lim_{x \to a} f(x)^{g(x)} \]

where \( f(x) \to \infty \) and \( g(x) \to 0 \), we first rewrite the expression using exponentials:

\[ \lim_{x \to a} f(x)^{g(x)} = \lim_{x \to a} e^{g(x) \ln f(x)}. \]

Now, the exponent

\[ \lim_{x \to a} g(x) \ln f(x) \]

takes the form \(0 \times \infty\), which is indeterminate. To resolve this, we rewrite it as

\[ \lim_{x \to a} g(x) \ln f(x) = \lim_{x \to a} \frac{\ln f(x)}{1/g(x)}. \]

Since \( \ln f(x) \to \infty \) and \( 1/g(x) \to \infty \), this is now an \(\infty/\infty\) form, which can be evaluated using L'Hôpital’s Rule, algebraic manipulation etc. Once this limit is computed, we exponentiate the result to obtain

\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} \frac{\ln f(x)}{1/g(x)}}. \]

Thus, the key idea in evaluating \(\infty^0\) limits is to transform them into an exponential form, convert the exponent into a standard indeterminate form (\(\infty/\infty\)), and apply standard techniques to determine the final value.

Example

Evaluate the limit

\[ \lim_{x \to \infty} x^{1/x}. \]

Solution. As \(x \to \infty\), the base \(x \to \infty\) and the exponent \(1/x \to 0\), creating the indeterminate form \(\infty^0\). Using the standard method, we rewrite the expression as an exponential:

\[ \lim_{x \to \infty} x^{1/x} = \lim_{x \to \infty} e^{\frac{\ln x}{x}}. \]

Now, evaluate the exponent inside the limit:

\[ \lim_{x \to \infty} \frac{\ln x}{x}. \]

Since \(\ln x \to \infty\) and \(x \to \infty\), this is an \(\infty/\infty\) form. Applying L’Hôpital’s Rule:

\[ \lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{( \ln x )'}{( x )'} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0. \]

Thus,

\[ \lim_{x \to \infty} e^{\frac{\ln x}{x}} = e^0 = 1. \]

Hence, the given limit evaluates to:

\[ \lim_{x \to \infty} x^{1/x} = 1. \]

Infinity raised to power zero: Special Case

A special case of the \(\infty^0\) form arises in the limit

\[ \lim_{x \to \infty} (2^x + 4^x)^{\frac{1}{2x+1}}. \]

As \( x \to \infty \), both \( 2^x \to \infty \) and \( 4^x \to \infty \), so the base tends to \(\infty\), while the exponent \( \frac{1}{2x+1} \to 0 \), creating the \(\infty^0\) form. To evaluate the limit, note that \( 4^x \) grows faster than \( 2^x \), so we factor out \( 4^x \) from the base:

\[ \lim_{x \to \infty} (2^x + 4^x)^{\frac{1}{2x+1}} = \lim_{x \to \infty} \left( 4^x \left( 1 + \frac{2^x}{4^x} \right) \right)^{\frac{1}{2x+1}} = \lim_{x \to \infty} \left( 4^x \left( 1 + 2^{-x} \right) \right)^{\frac{1}{2x+1}}. \]

Applying exponent properties,

\[ \lim_{x \to \infty} 4^{\frac{x}{2x+1}} \cdot \left(1 + 2^{-x}\right)^{\frac{1}{2x+1}}. \]

Since \(2^{-x} \to 0\), \(1 + 2^{-x} \) approaches \(1\) and the power \(\frac{1}{2x+1}\) approaches \(0\) as \(x \to \infty\), reducing the limit to

\[ \lim_{x \to \infty} 4^{\frac{x}{2x+1}}. \]

Now, simplifying the exponent of base \(4\),

\[ \frac{x}{2x+1} = \frac{x}{2x(1 + \frac{1}{2x})} = \frac{1}{2(1 + \frac{1}{2x})}. \]

Since \(\frac{1}{2x} \to 0\), we obtain \(\frac{1}{2(1+0)} = \frac{1}{2}\), giving

\[ \lim_{x \to \infty} 4^{\frac{1}{2}} = 2. \]

Thus,

\[ \lim_{x \to \infty} (2^x + 4^x)^{\frac{1}{2x+1}} = 2. \]

Consider another example:

Example

Evaluate the limit

\[ \lim_{x \to 0} \left( 1^{\csc^2 x} + 2^{\csc^2 x} + 3^{\csc^2 x} + \dots + 10^{\csc^2 x} \right)^{\sin^2 x}. \]

Solution. As \( x \to 0 \), we observe that \( \csc^2 x = \frac{1}{\sin^2 x} \to \infty \), which causes terms like \( k^{\csc^2 x} \) to grow rapidly when \( k > 1 \), while the first term \( 1^{\csc^2 x} = 1 \) remains constant. Thus, the dominant term in the sum is \( 10^{\csc^2 x} \), which grows the fastest. Factoring out \( 10^{\csc^2 x} \):

\[ \begin{align} \lim_{x \to 0} \left( 1^{\csc^2 x} + 2^{\csc^2 x} + \dots + 10^{\csc^2 x} \right)^{\sin^2 x} &= \lim_{x \to 0} \left( 10^{\csc^2 x} \left( \frac{1^{\csc^2 x}}{10^{\csc^2 x}} + \frac{2^{\csc^2 x}}{10^{\csc^2 x}} + \dots + 1 \right) \right)^{\sin^2 x} \\ &= \lim_{x \to 0} \left( 10^{\csc^2 x} \right)^{\sin^2 x} \cdot \left( \left(\frac{1}{10}\right)^{\csc^2 x} + \left(\frac{2}{10}\right)^{\csc^2 x} + \left(\frac{3}{10}\right)^{\csc^2 x} + \dots + 1 \right)^{\sin^2 x}. \end{align} \]

Since \( \csc^2 x = \frac{1}{\sin^2 x} \), we rewrite the dominant term:

\[ \lim_{x \to 0} \left( 10^{\csc^2 x} \right)^{\sin^2 x} = \lim_{x \to 0} 10 = 10. \]

For the remaining factor,

\[ \lim_{x \to 0} \left( \left(\frac{1}{10}\right)^{\csc^2 x} + \left(\frac{2}{10}\right)^{\csc^2 x} + \left(\frac{3}{10}\right)^{\csc^2 x} + \dots + 1 \right)^{\sin^2 x} = 1 \]

Since the last term remains constant at 1, while all other terms approach 0 because their bases are less than 1 and their exponents tend to infinity, the entire sum inside the parentheses converges to 1. Raising this expression to \(\sin^2 x\), which approaches 0, ensures that the limit remains 1.

Thus, the final result is:

\[ \lim_{x \to 0} \left( 1^{\csc^2 x} + 2^{\csc^2 x} + \dots + 10^{\csc^2 x} \right)^{\sin^2 x} = 10. \]