Limit Toolbox Limit Toolbox
Introduction
For the next few lessons, we will be building our limit toolbox. The first tools we
will learn are the following:
The Constant Multiple Rule
\lim_{x \to a} [ c \cdot f(x) ] = c \cdot \lim_{x \to a} f(x)
The limit of a constant and a function is equal to the product of the constant and the limit of the function.
\lim_{x \to a} [ f(x) + g(x) ] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)
The limit of two summed functions is equal to the sum of their limits.
\lim_{x \to a} [ f(x) \cdot g(x) ] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)
The limit of two multiplied functions is equal to the product of their limits.
\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \text{, if } \lim_{x \to a} g(x) \neq 0
The limit of two divided functions is equal to the quotient of their limits (as long as the denominator is not
0
).
Proving the Constant Multiple Rule
\lim_{x \to a} [ c \cdot f(x) ] = L
\forall \epsilon \gt 0, \exists \delta \gt 0 \text{ where if } (|x-a| \lt \delta) \text{, then }(|c \cdot f(x)-L| \lt |c| \cdot \epsilon)
(\frac{|c \cdot f(x)-L|}{|c|} \lt \epsilon)
(|f(x)-\frac{L}{c}| \lt \epsilon)
\lim_{x \to a} f(x) = \frac{L}{c}
c \cdot \lim_{x \to a} f(x) = L
\lim_{x \to a} [ c \cdot f(x) ] = c \cdot \lim_{x \to a} f(x)
Proving the Sum Rule
\lim_{x \to a} [ f(x) + g(x) ] = L + M
\forall \epsilon \gt 0, \exists \delta \gt 0 \text{ where if } |x - a| \lt \delta \text{, then } |f(x) + g(x) - (L + M)| \lt \epsilon
|f(x) - L + g(x) - M| \leq |f(x) - L| + |g(x) - M|
\text{Choose } \delta_1 \text{ such that } |f(x) - L| \lt \frac{\epsilon}{2}
\text{Choose } \delta_2 \text{ such that } |g(x) - M| \lt \frac{\epsilon}{2}
\delta = \text{ the minimum of } \delta_1 \text{ and } \delta_2
|f(x) - L| + |g(x) - M| \lt \epsilon
\lim_{x \to a} [ f(x) + g(x) ] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)
Proving the Product Rule
\lim_{x \to a} [ f(x) \cdot g(x) ] = L \cdot M
\forall \epsilon \gt 0, \exists \delta \gt 0 \text{ where if } |x - a| \lt \delta \text{, then } |f(x)g(x) - LM| \lt \epsilon
|f(x)g(x) - LM| = |f(x)g(x) - Lg(x) + Lg(x) - LM| \leq |g(x)||f(x) - L| + |L||g(x) - M|
\text{Bound } |g(x)| \text{ near } a: \exists \delta_1 \text{ where } |g(x)| \lt |M| + 1
\text{Choose } \delta_2 \text{ for } \frac{\epsilon}{2(|M| + 1)} \text{ in } \lim_{x \to a} f(x) = L
\text{Choose } \delta_3 \text{ for } \frac{\epsilon}{2|L|} \text{ in } \lim_{x \to a} g(x) = M
\delta = \min(\delta_1, \delta_2, \delta_3) \text{ ensures } |f(x) - L| + |g(x) - M| \lt \epsilon
\lim_{x \to a} [ f(x) \cdot g(x) ] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)
Proving the Quotient Rule
\lim_{x \to a} \left( \frac{f(x)}{g(x)} \right) = \frac{L}{M} \text{ (assuming } M \neq 0\text{)}
\text{First show } \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M}
\forall \epsilon \gt 0, \exists \delta \gt 0 \text{ where } |x - a| \lt \delta \implies \left| \frac{1}{g(x)} - \frac{1}{M} \right| \lt \epsilon
\left| \frac{M - g(x)}{M g(x)} \right| = \frac{|g(x) - M|}{|M||g(x)|}
\text{Ensure } |g(x)| \gt \frac{|M|}{2} \text{ by choosing } \delta_1 \text{ where } |g(x) - M| \lt \frac{|M|}{2}
\text{Then } \frac{1}{|g(x)|} \lt \frac{2}{|M|}
\text{Choose } \delta_2 \text{ for } \frac{\epsilon |M|^2}{2} \text{ in } \lim_{x \to a} g(x) = M
\delta = \min(\delta_1, \delta_2) \implies \frac{|g(x) - M|}{|M||g(x)|} \lt \epsilon
\lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M}
\text{Apply Product Rule: } \lim_{x \to a} \frac{f(x)}{g(x)} = L \cdot \frac{1}{M} = \frac{L}{M}
\lim_{x \to a} \left( \frac{f(x)}{g(x)} \right) = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}