Question

(a). Prove Theorem 4, part 3.

(b). Prove Theorem 4, part 5.

Short answer

a). It is proved that the function \(cf\) is continuous at \(a\).

b). It is proved that the function \(\frac{f}{g}\) is continuous at \(a\).

Step-by-step solution

The statement in Theorem 4

Theorem 4states that when\(f\)and\(g\)arecontinuous at a number\(a\)and\(c\)is aconstant, then the following function is also continuous at\(a\):

1. \(f + g\)
2. \(f - g\)
3. \(cf\)
4. \(fg\)
5. \(\frac{f}{g},\,\,{\mathop{\rm if}\nolimits} \,\,g\left( a \right) \ne 0\).

Prove that the function \(cf\) and \(\frac{f}{g}\) are continuous at a

a)

The function \(f\) is continuous at \(a\) such that \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\).

Use the Constant Multiple Law of Limits to prove that\(cf\)is continuous at\(a\)as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to a} \left( {cf} \right)\left( x \right)& = \mathop {\lim }\limits_{x \to a} cf\left( x \right)\\ &= c\mathop {\lim }\limits_{x \to a} f\left( x \right)\\ &= cf\left( a \right)\\ &= \left( {cf} \right)\left( a \right)\end{aligned}\)

Thus, it is proved that the function\(cf\)is continuous at\(a\).

b)

The function\(f\)and\(g\)are continuous at\(a\)such that\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\)and\(\mathop {\lim }\limits_{x \to a} g\left( x \right) = g\left( a \right)\).

Use the Quotient Law of Limits because\(g\left( a \right) \ne 0\)to prove that\(\frac{f}{g}\)is continuous at\(a\)as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to a} \left( {\frac{f}{g}} \right)\left( x \right) &= \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}}\\ &= \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\\ &= \frac{{f\left( a \right)}}{{g\left( a \right)}}\\ &= \frac{f}{g}\left( a \right)\end{aligned}\)

Thus, it is proved that the function\(\frac{f}{g}\)is continuous at\(a\).