Question

Find functions \(f\) and \(g\) where \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} g\left( x \right) = \infty \) and

(a) \(\mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{g\left( x \right)}} = 7\) (b) \(\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right) = 7\)

Short answer

(a) \(f\left( x \right) = \frac{7}{{{x^2}}}\) and \(g\left( x \right) = \frac{1}{{{x^2}}}\)

(b)\(f\left( x \right) = \frac{1}{{{x^2}}} + 7\) and \(g\left( x \right) = \frac{1}{{{x^2}}}\)

Step-by-step solution

(a) Step 1: Find the functions \(f\) and \(g\)

Consider the limit \(\mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{g\left( x \right)}} = 7\).

As the functions \(f\) and \(g\) whose limits are \(\infty \) as \(x \to 0\) and the quotients of the functions has a limit of \(7\) as \(x \to 0\).

Now write the pair of functions.

\(f\left( x \right) = \frac{7}{{{x^2}}}\)and \(g\left( x \right) = \frac{1}{{{x^2}}}\)

The limits of the function are shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} g\left( x \right)\\ = \infty \end{array}\)

Now solve the limit. \(\mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{g\left( x \right)}}\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{7}{{{x^2}}}}}{{\frac{1}{{{x^2}}}}}\\ = \mathop {\lim }\limits_{x \to 0} 7\\ = 7\end{array}\)

Thus, \(f\left( x \right) = \frac{7}{{{x^2}}}\) and \(g\left( x \right) = \frac{1}{{{x^2}}}\).

(b) Step 2: Find the functions \(f\) and \(g\)

Consider the limit \(\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right) = 7\).

As the functions \(f\) and \(g\) whose limits are \(\infty \) as \(x \to 0\) and difference of the functions has a limit of \(7\) as \(x \to 0\).

Now write the pair of functions.

\(f\left( x \right) = \frac{1}{{{x^2}}} + 7\)and \(g\left( x \right) = \frac{1}{{{x^2}}}\)

The limits of the function are shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} g\left( x \right)\\ = \infty \end{array}\)

Now solve the limit. \(\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right)\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\left( {\frac{1}{{{x^2}}} + 7} \right) - \frac{1}{{{x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to 0} 7\\ = 7\end{array}\)

Thus, \(f\left( x \right) = \frac{1}{{{x^2}}} + 7\) and \(g\left( x \right) = \frac{1}{{{x^2}}}\).