Limit Laws
- $\lim\limits_{x \to a}\ \left( f(x) + g(x) \right) = \ ?$
Answer:
$\lim\limits_{x \to a}\ \left( f(x) + g(x) \right) = \lim\limits_{x \to a}\ f(x) + \lim\limits_{x \to a}\ g(x).$ - $\lim\limits_{x \to a}\ f(x)\ g(x) = \ ?$
Answer:
$$\lim\limits_{x \to a}\ \left(f(x) \ g(x) \right) = \left(\lim\limits_{x \to a}\ f(x)\right) \left(\lim\limits_{x \to a}\ g(x)\right).$$ - $\lim\limits_{x \to a}\ \dfrac{f(x) }{ g(x) } = \ ?$
Answer:
$$\lim\limits_{x \to a}\ \dfrac{f(x) }{ g(x) } = \dfrac{\lim\limits_{x \to a}\ f(x) }{\lim\limits_{x \to a}\ g(x) }.$$ - $\lim\limits_{x \to a}\ f \left( g(x) \right) = \ ?$
Answer:
$$\lim\limits_{x \to a}\ f \left( g(x) \right) = f \left( \lim\limits_{x \to a}g(x) \right),$$ assuming $f$ is a continuous function. - $\lim\limits_{x \to a}\ 17 = \ ?$
Answer:
$$\lim\limits_{x \to a}\ 17 = 17.$$ - $\lim\limits_{x \to a}\ \left( f(x) \right)^2 = \ ?$
Answer:
$$\lim\limits_{x \to a}\ \left( f(x) \right)^2 = \left( \lim\limits_{x \to a}f(x) \right)^2.$$ - $\lim\limits_{x \to a}\ \left( f(x) \right)^n = \ ?$
Answer:
$$\lim\limits_{x \to a}\ \left( f(x) \right)^n = \left( \lim\limits_{x \to a} f(x) \right)^n .$$ - $\lim\limits_{x \to a}\ (7x - 2)^3 = \ ?$
Answer:
\begin{eqnarray} \lim\limits_{x \to a}\ (7x - 2)^3 &=& \left(\lim\limits_{x \to a}\ (7x - 2) \right)^3 \\ \ &=& (7a - 2)^3 . \end{eqnarray} - $\lim\limits_{x \to 0}\ \sqrt{x+4} = \ ?$
Answer:
Since the square root function is continuous, \begin{eqnarray} \lim\limits_{x \to 0}\ \sqrt{x+4} &=& \sqrt{\lim\limits_{x \to 0}\ (x+4)} \\ \ &=& \sqrt {4} \\ \ &=& 2. \end{eqnarray} - $\lim\limits_{x \to\ -7}\ \sqrt{x+4} = \ ?$
Answer:
\begin{eqnarray} \lim\limits_{x \to\ -7}\ \sqrt{x+4} &=& \sqrt{-7 + 4} \\ \ &=& \sqrt{-3} \\ \ &=& i \sqrt{3}, \end{eqnarray} where $i = \sqrt{-1}$ is called the "imaginary number."