Answer to Question #309604 in Calculus for Shane

Question

Activity in Limit Theorems

Directions: Assume the following.

1. lim f(x) = 3/4

x→c

2. lim g(x) = 12

x→c

3. lim h(x) = -3

x→c

Answer 1

Solution

Given that

$\mathop {\lim }\limits_{x \to c} f\left( x \right) = \frac{3}{4}\\$

$\mathop {\lim }\limits_{x \to c} g\left( x \right) = 12\\$

$\mathop {\lim }\limits_{x \to c} h\left( x \right) = - 3$

We can solve the following different limits which will be using the above-given values.

Example (1)

$\mathop {\lim }\limits_{x \to c} \left[ {4 \cdot f\left( x \right)} \right] = 4\mathop { \cdot \lim }\limits_{x \to c} f\left( x \right) = 4 \cdot \frac{3}{4} = 3\$

Example (2)

$\mathop {\lim }\limits_{x \to c} \left[ {g\left( x \right) - h\left( x \right)} \right] = \mathop {\lim }\limits_{x \to c} g\left( x \right) - \mathop {\lim }\limits_{x \to c} h\left( x \right) = 12 - \left( { - 3} \right) = 12 + 3 = 15\$

Example (3)

$\mathop {\lim }\limits_{x \to c} \sqrt {12 \cdot f\left( x \right)} = \sqrt {12 \cdot \mathop {\lim }\limits_{x \to c} f\left( x \right)} = \sqrt {12 \cdot \frac{3}{4}} = \sqrt {3 \cdot 3} = \sqrt 9 = 3\$

Example (4)

$\mathop {\lim }\limits_{x \to c} \left( {\frac{{g\left( x \right) + h\left( x \right)}}{{f\left( x \right)}}} \right) = \frac{{\mathop {\lim }\limits_{x \to c} g\left( x \right) + \mathop {\lim }\limits_{x \to c} h\left( x \right)}}{{\mathop {\lim }\limits_{x \to c} f\left( x \right)}} = \frac{{12 + \left( { - 3} \right)}}{{\frac{3}{4}}} = 9 \times \frac{4}{3} = 12\$

Example (5)

$\mathop {\lim }\limits_{x \to c} \left( {f\left( x \right) + h\left( x \right)} \right) = \mathop {\lim }\limits_{x \to c} f\left( x \right) + \mathop {\lim }\limits_{x \to c} h\left( x \right) = \frac{3}{4} + \left( { - 3} \right) = \frac{{3 - 12}}{4} = - \frac{9}{4}\$