INDETERMINATE FORMS
These are the seven standard indeterminate forms that occur in limit problems. Each one needs special techniques such as L'Hôpital's Rule or algebraic manipulation.
L'Hospital's Rule
If \( f(x) \) and \( g(x) \) are two functions such that their derivatives \( f'(x) \) and \( g'(x) \) are continuous at \( x = a \) and \( f(a) = g(a) = 0 \), then
\[ \lim_{x \to a} \frac{f(x)}{g(x)} \;=\; \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{if } g'(a) \neq 0. \]
⚠️ WARNING !!
L'Hospital's Rule can be used only for the indeterminate forms \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \).
For the other indeterminate forms, we must first manipulate or transform the function so that it becomes \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hospital's Rule.
PROBLEM 1
Evaluate \( \displaystyle \lim_{x\to0} \frac{\tan x}{x} \)
\[ \lim_{x\to0} \frac{\tan x}{x} \quad \left( \frac{0}{0} \right) \]
\[ = \lim_{x\to0} \frac{\sec^{2} x}{1} \] (Applying L'Hospital's Rule)
\[ = \sec^{2} 0 \]
\[ = 1 \]
PROBLEM 2
Evaluate \( \displaystyle \lim_{x\to0} \frac{x e^{x} - (1+x)\log(1+x)}{x^{2}} \)
Given
\[ \lim_{x\to0} \frac{x e^{x} - (1+x)\log(1+x)}{x^{2}} \quad \left( \frac{0}{0} \right) \]
\[ = \lim_{x\to0} \frac{x e^{x} + e^{x}\cdot 1 - \left[(1+x)\frac{1}{1+x} + \log(1+x)\cdot 1\right]} {2x} \]
(Applying L'Hospital's Rule)
\[ = \lim_{x\to0} \frac{x e^{x} + e^{x}\cdot 1 + e^{x} - 0 - \frac{1}{1+x}}{2} \]
\[ = \frac{0 + 1 + 1 - \frac{1}{1+0}}{2} \]
\[ = \frac{1}{2} \]
PROBLEM 3
Evaluate \( \displaystyle \lim_{x\to0} \frac{\log(\sin x)}{\cot x} \)
Given
\[ \lim_{x\to0} \frac{\log(\sin x)}{\cot x} \quad \left( \frac{\infty}{\infty} \right) \]
\[ = \lim_{x\to0} \frac{\frac{1}{\sin x}\cdot \cos x}{-\csc^{2} x} \] (Applying L'Hospital's Rule)
AFTER APPLYING L'HOSPITAL'S RULE, ALWAYS SIMPLIFY…
\[ = -\lim_{x\to0} \frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin^{2} x}} \]
\[ = - \lim_{x\to0} \cos x \,\sin x \]
\[ = -1 \times 0 \]
\[ = 0 \]
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