When we have limits of the form \( [f(x)]^{g(x)} \) that give the indeterminate types \( 1^{\infty},\; 0^{0},\; \infty^{0} \), we always convert them using logarithms.
The general limit looks like: \[ \lim_{x \to a} \left( f(x) \right)^{g(x)} \] and we follow these standard steps:
Write \( L = \displaystyle \lim\limits_{x \to a} \left( f(x) \right)^{g(x)} \).
Use the property \( \ln(a^{b}) = b \ln a \) to write \[ \ln L = \lim_{x \to a} g(x)\,\ln\big( f(x) \big). \]
Rewrite
\( g(x)\,\ln(f(x)) \)
so that the limit becomes a quotient of the form
\( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
(For example, turn a product into a fraction.)
Use L'Hospital's Rule on the limit for \( \ln L \) to find a finite value \( K \) such that \( \ln L = K \).
If \( \ln L = K \), then \( L = e^{K} \).
Evaluate \( \displaystyle \lim_{x\to0} x^{\,x} \)
Given
\[ \lim_{x\to0} x^{\,x} \quad (0^{0}) \]
Let \( L = \displaystyle \lim_{x\to0} x^{\,x} \).
Taking natural log on both sides:
\[ \ln L = \ln\!\left( \lim_{x\to0} x^{\,x} \right) \] \[ \ln L = \lim_{x\to0} \ln\!\left( x^{\,x} \right) \] \[ \ln L = \lim_{x\to0} x \ln x \]
\[ \ln L = \lim_{x\to0} x \ln x \quad (0 \times -\infty) \]
Rewrite as a quotient: \[ \ln L = \lim_{x\to0} \frac{\ln x}{\frac{1}{x}} \quad \left( \frac{-\infty}{\infty} \right) \]
(Applying L'Hospital's Rule) \[ \ln L = \lim_{x\to0} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}} \]
\[ \ln L = \lim_{x\to0} (-x) \]
\[ \ln L = 0 \]
\[ L = e^{0} \]
\[ = 1 \]
Evaluate \( \displaystyle \lim_{x\to0} (1+x)^{\,\frac{1}{x}} \)
Given
\[ \lim_{x\to0} (1+x)^{\,\frac{1}{x}} \quad (1^{\infty}) \]
Let \( L = \displaystyle \lim_{x\to0} (1+x)^{\,\frac{1}{x}} \).
Taking natural log on both sides:
\[ \ln L = \ln\!\left( \lim_{x\to0} (1+x)^{\,\frac{1}{x}} \right) \] \[ \ln L = \lim_{x\to0} \ln\!\left( (1+x)^{\,\frac{1}{x}} \right) \] \[ \ln L = \lim_{x\to0} \frac{1}{x}\,\ln(1+x) \]
\[ \ln L = \lim_{x\to0} \frac{1}{x}\,\ln(1+x) \quad (\infty \times 0) \]
Rewrite as a quotient: \[ \ln L = \lim_{x\to0} \frac{\ln(1+x)}{x} \quad \left( \frac{0}{0} \right) \]
(Applying L'Hospital's Rule) \[ \ln L = \lim_{x\to0} \frac{\frac{1}{1+x}}{1} \]
\[ \ln L = \frac{1}{1+0} = 1 \]
\[ L = e^{1} \]
\[ = e \]
Evaluate \( \displaystyle \lim_{x\to0} \big( \cot x \big)^{\sin 2x} \)
Given
\[ \lim_{x\to0} \big( \cot x \big)^{\sin 2x} \quad (1^{\infty}) \]
Let \( L = \displaystyle \lim_{x\to0} \big( \cot x \big)^{\sin 2x} \).
Taking natural log on both sides:
\[ \ln L = \ln\!\left( \lim_{x\to0} \big( \cot x \big)^{\sin 2x} \right) \] \[ \ln L = \lim_{x\to0} \ln\!\left( \big( \cot x \big)^{\sin 2x} \right) \] \[ \ln L = \lim_{x\to0} \sin 2x \,\ln(\cot x) \]
\[ \ln L = \lim_{x\to0} \sin 2x \,\ln(\cot x) \quad (0 \cdot \infty) \]
Rewrite as a quotient: \[ \ln L = \lim_{x\to0} \frac{\ln(\cot x)}{\csc 2x} \quad \left( \frac{\infty}{\infty} \right) \]
(Applying L'Hospital's Rule) \[ \ln L = \lim_{x\to0} \frac{\frac{1}{\cot x}\,(-\csc^{2}x)}{-2\,\csc 2x \,\cot 2x} \]
\[ \ln L = \frac{1}{2} \lim_{x\to0} \frac{1}{\cot x}\,\csc^{2}x \cdot \frac{1}{\csc 2x \,\cot 2x} \]
\[ \ln L = \frac{1}{2} \lim_{x\to0} \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^{2}x} \cdot \sin 2x \,\tan 2x \]
\[ \ln L = \frac{1}{2} \lim_{x\to0} \frac{1}{\cos x \,\sin x} \cdot 2 \sin x \cos x \,\tan 2x \]
\[ \ln L = \lim_{x\to0} \tan 2x \]
\[ \ln L = 0 \]
\[ L = e^{0} \]
\[ = 1 \]