Use the Table of Values of F to Estimate the Limit

Use the table of values of f to estimate the limit.
- $\text { Let } f(x) =\frac{x^{2}+7 x+12}{x^{2}+5 x+4} \text {, find } \lim _{x \rightarrow 4} f(x)$
$\begin{array}{c|l|l|l|l|l|l}\mathrm{x} & -4.1 & -4.01 & -4.001 & -3.999 & -3.99 & -3.9 \\\hline \mathrm{f}(\mathrm{x}) & & & & & &\end{array}$

A)
$\begin{array}{c|cccccc}\mathrm{x} & -4.1 & -4.01 & -4.001 & -3.999 & -3.99 & -3.9 \\\hline \mathrm{f}(\mathrm{x}) & 0.2548 & 0.2355 & 0.2336 & 0.2331 & 0.2311 & 0.2103\end{array} \text { limit }=0.2333$

B)
$\begin{array}{c|cccccc}\mathrm{x} & -4.1 & -4.01 & -4.001 & -3.999 & -3.99 & -3.9 \\\hline \mathrm{f}(\mathrm{x}) & 0.4548 & 0.4355 & 0.4336 & 0.4331 & 0.4311 & 0.4103\end{array} \text {; limit }=0.4333$

C)
$\begin{array}{c|cccccc}\mathrm{x} & -4.1 & -4.01 & -4.001 & -3.999 & -3.99 & -3.9 \\\hline \mathrm{f}(\mathrm{x}) & 0.3548 & 0.3355 & 0.3336 & 0.3331 & 0.3311 & 0.3103\end{array} \text {; limit }=0.3333$

D)
$\begin{array}{c|cccccc}\mathrm{x} & -4.1 & -4.01 & -4.001 & -3.999 & -3.99 & -3.9 \\\hline \mathrm{f}(\mathrm{x}) & 1.4082 & 1.4008 & 1.4001 & 1.3999 & 1.3992 & 1.3922\end{array} \text {; limit }=1.4$