Question

If \(4x - 9 \le f\left( x \right) \le {x^2} - 4x + 7\) for \(x \ge 0\), find \(\mathop {\lim }\limits_{x \to 4} f\left( x \right)\).

Short answer

The value of the limit is \(\mathop {\lim }\limits_{x \to 4} f\left( x \right) = 7\).

Step-by-step solution

The Squeeze theorem

When \(f\left( x \right) \le g\left( x \right) \le h\left( x \right)\), if\(x\) is near \(a\) (except possibly at \(a\)) and

\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} h\left( x \right) = L\) then \(\mathop {\lim }\limits_{x \to a} g\left( x \right) = L\).

Determine \(\mathop {\lim }\limits_{x \to 4} f\left( x \right)\)

Consider that \(\left( {4x - 9} \right) \le f\left( x \right) \le {x^2} - 4x + 7\).

Obtain the limit\(\mathop {\lim }\limits_{x \to 4} f\left( x \right)\) as shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 4} \left( {4x - 9} \right) &=& 4\left( 4 \right) - 9\\ &=& 16 - 9\\ &=& 7\end{array}\)

And,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 4} \left( {{x^2} - 4x + 7} \right) &=& {4^2} - 4\left( 4 \right) + 7\\ &=& 7\end{array}\)

It follows that \(\mathop {\lim }\limits_{x \to 4} f\left( x \right) = 7\) according to the Squeeze theorem because \(\left( {4x - 9} \right) \le f\left( x \right) \le {x^2} - 4x + 7\) for any \(x \ge 0\).

Thus, the value of the limit is \(\mathop {\lim }\limits_{x \to 4} f\left( x \right) = 7\).