Master the technique for solving limits with indeterminate power forms (1^infinity, 0^0, and infinity^0). This method involves using the natural logarithm to convert the problem into a product, which can then be solved with L'Hopital's Rule.
The three indeterminate power forms are: \(1^\infty\), \(0^0\), and \(\infty^0\). Direct substitution into these forms does not yield a determinate answer.
What is the crucial first step for evaluating a limit with an indeterminate power, \(\lim_{x \to c} [f(x)]^{g(x)}\)?
Let \(y = [f(x)]^{g(x)}\), then take the natural logarithm of both sides to get \(\ln(y) = g(x) \ln(f(x))\). This transforms the power into a product.
After finding the limit of the logarithm, \(\lim_{x \to c} \ln(y) = L\), what is the final step?
The limit of the original function \(y\) is \(e^L\). This is because if \(\lim \ln(y) = L\), then \(\lim y = \lim e^{\ln y} = e^{\lim \ln y} = e^L\).
Identify the indeterminate form of the limit that defines \(e\): $$ \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x $$
As \(x \to \infty\), the base \((1 + \frac{1}{x}) \to 1\) and the exponent \(x \to \infty\). This is the indeterminate form \(1^\infty\).
To solve \(\lim_{x \to \infty} (1 + \frac{1}{x})^x\), we set \(y = (1 + \frac{1}{x})^x\) and find \(\lim_{x \to \infty} \ln(y)\). What does \(\ln(y)\) become?
\(\ln(y) = \ln\left((1 + \frac{1}{x})^x\right) = x \ln(1 + \frac{1}{x})\). This is a \(\infty \cdot 0\) form, which must be converted to a quotient.
Evaluate \(\lim_{x \to \infty} x \ln(1 + \frac{1}{x})\) by first converting it to a quotient.
Rewrite as \(\lim_{x \to \infty} \frac{\ln(1 + 1/x)}{1/x}\). This is a \(\frac{0}{0}\) form. Using L'Hôpital's Rule gives \(\lim_{x \to \infty} \frac{\frac{1}{1+1/x} \cdot (-1/x^2)}{-1/x^2} = \lim_{x \to \infty} \frac{1}{1+1/x} = 1\).
Given that \(\lim_{x \to \infty} \ln\left((1 + \frac{1}{x})^x\right) = 1\), what is the value of the original limit?
Since the limit of the logarithm is \(1\), the original limit is \(e^1 = e\).
What is the indeterminate form of \(\lim_{x \to 0^+} x^x\)?
As \(x \to 0^+\), both the base and the exponent approach \(0\). This is the indeterminate form \(0^0\).
Evaluate \(\lim_{x \to 0^+} x^x\).
Let \(y=x^x\), so \(\ln(y) = x \ln(x)\). The limit \(\lim_{x \to 0^+} x \ln(x) = 0\). Therefore, the original limit is \(e^0 = 1\).
What is the indeterminate form of \(\lim_{x \to \infty} x^{1/x}\)?
As \(x \to \infty\), the base \(x \to \infty\) and the exponent \(1/x \to 0\). This is the indeterminate form \(\infty^0\).
Evaluate the limit \(\lim_{x \to \infty} x^{1/x}\).
Let \(y = x^{1/x}\), so \(\ln(y) = \frac{\ln(x)}{x}\). This limit is an \(\frac{\infty}{\infty}\) form. Using L'Hôpital's Rule: \(\lim_{x \to \infty} \frac{1/x}{1} = 0\). The original limit is \(e^0 = 1\).
Why isn't \(0^\infty\) an indeterminate form?
This form is determinate. A number very close to zero raised to a very large power becomes even closer to zero. The limit is \(0\).
Why isn't \(\infty^\infty\) an indeterminate form?
This form is determinate. A very large number raised to a very large power results in an even larger number. The limit is \(\infty\).
