Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 35

Suppose $4x - 9 \leq f(x) \leq x^2 - 4x + 7$ for $x \geq 0$, find $\lim \limits_{x \to 4} f(x)$


$
\begin{equation}
\begin{aligned}

& \text{Using Squeeze Theorem}\\
\phantom{x}& && \lim \limits_{x \to 4} (4x-9) \leq \lim \limits_{x \to 4} f(x) \leq \lim \limits_{x \to 4} (x^2-4x+7)\\
\phantom{x}& && 4(4)-9 \leq \lim \limits_{x \to 4} f(x) \leq [(4)^2 - 4(4) +7]\\
\phantom{x}& && 7 \leq \lim \limits_{x \to 4} f(x) \leq 7\\

& \text{We have,}\\
\phantom{x}& && \lim \limits_{x \to 4} (4x-9) = \lim \limits_{x \to 4} (x^2-4x+7)\\

& \text{Therefore,}\\
\phantom{x}& && \lim \limits_{x \to 4} f(x) = 7

\end{aligned}
\end{equation}
$