Limits: (lesson 1 of 5)
Properties of Limits
If both sides exist, it can be proved that:
lim x → a ( f ( x ) + g ( x ) ) = lim x → a f ( x ) + lim x → a g ( x ) lim x → a c ⋅ f ( x ) = c ⋅ lim x → a g ( x ) lim x → a ( f ( x ) ⋅ g ( x ) ) = lim x → a f ( x ) ⋅ lim x → a g ( x ) lim x → a f ( x ) g ( x ) = lim x → a f ( x ) lim x → a g ( x ) ( if lim x → a g ( x ) ≠ 0 ) lim x → a f ( x ) p = ( lim x → a f ( x ) ) p lim x → a ∣ f ( x ) ∣ = ∣ lim x → a f ( x ) ∣
\begin{aligned}
&\mathop {\lim }\limits_{x \to a} (f(x) + g(x)) = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x) \\ \\
&\mathop {\lim }\limits_{x \to a} c \cdot f(x) = c \cdot \mathop {\lim }\limits_{x \to a} g(x) \\ \\
&\mathop {\lim }\limits_{x \to a} (f(x) \cdot g(x)) = \mathop {\lim }\limits_{x \to a} f(x) \cdot \mathop {\lim }\limits_{x \to a} g(x) \\ \\
&\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}} \ (\text{if} \ \mathop {\lim }\limits_{x \to a} g(x) \ne 0) \\ \\
&\mathop {\lim }\limits_{x \to a} f{(x)^p} = {\left( {\mathop {\lim }\limits_{x \to a} f(x)} \right)^p} \\ \\
&\mathop {\lim }\limits_{x \to a} \left| {f(x)} \right| = \left| {\mathop {\lim }\limits_{x \to a} f(x)} \right|
\end{aligned}
x → a lim ( f ( x ) + g ( x )) = x → a lim f ( x ) + x → a lim g ( x ) x → a lim c ⋅ f ( x ) = c ⋅ x → a lim g ( x ) x → a lim ( f ( x ) ⋅ g ( x )) = x → a lim f ( x ) ⋅ x → a lim g ( x ) x → a lim g ( x ) f ( x ) = x → a lim g ( x ) x → a lim f ( x ) ( if x → a lim g ( x ) = 0 ) x → a lim f ( x ) p = ( x → a lim f ( x ) ) p x → a lim ∣ f ( x ) ∣ = x → a lim f ( x )
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Archimedes of Syracus
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Bertrand Russell
