Note: for the indeterminate form \( 0 \times \infty \) we cannot apply L'Hospital's Rule directly.
So we use the trick on one of the functions: \[ f(x) = \frac{1}{\frac{1}{f(x)}} , \] and rewrite the product as a quotient of type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
Examples:
\[ x = \frac{1}{\frac{1}{x}} \]
\[ \sin 2x = \frac{1}{\frac{1}{\sin 2x}} = \frac{1}{\csc 2x} \]
PROBLEM 7
Evaluate \( \displaystyle \lim_{x\to0} x \log x \)
Given
\[ \lim_{x\to0} x \log x \quad \left(0 \times -\infty \right) \]
\[ = \lim_{x\to0} \frac{\log x}{\frac{1}{x}} \quad \left( \frac{-\infty}{\infty} \right) \]
(rewrite \(x = \frac{1}{\frac{1}{x}}\) to convert the product into a quotient)
\[ = \lim_{x\to0} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}} \]
(Applying L'Hospital's Rule)
\[ = \lim_{x\to0} (-x) \]
\[ = 0 \]
(since \( \frac{A/B}{C/D} = \frac{A}{B} \times \frac{D}{C} \))
PROBLEM 8
Evaluate \( \displaystyle \lim_{x\to0} x \log(\tan x) \)
Given
\[ \lim_{x\to0} x \log(\tan x) \quad \left( 0 \times -\infty \right) \]
\[ = \lim_{x\to0} \frac{\log(\tan x)}{\frac{1}{x}} \quad \left( \frac{-\infty}{\infty} \right) \]
\[ = \lim_{x\to0} \frac{\frac{1}{\tan x}\,\sec^{2} x}{-\frac{1}{x^{2}}} \]
(Applying L'Hospital's Rule)
Simplifying
\[ = \lim_{x\to0} \frac{1}{\tan x} \times \sec^{2} x \times \frac{x^{2}}{-1} \]
\[ = - \lim_{x\to0} \frac{x^{2}}{\tan x} \,\sec^{2} x \]
\[ = - \lim_{x\to0} x \times \frac{x}{\tan x} \times \sec^{2} x \]
\[ = -\,0 \times 1 \times 1 \]
\[ = 0 \]
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