Use the Table of Values of F to Estimate the Limit

Multiple Choice

Use the table of values of f to estimate the limit.
- $\text { Let } f(x) =x^{2}-5 \text {, find } \lim _{x \rightarrow \theta} f(x)$
$\begin{array}{c|c|c|c|c|c|c}\mathrm{x} & -0.1 & -0.01 & -0.001 & 0.001 & 0.01 & 0.1 \\\hline \mathrm{f}(\mathrm{x}) & & & & & &\end{array}$

A)
$\begin{array}{c|llllcc}\mathrm{x} & -0.1 & -0.01 & -0.001 & 0.001 & 0.01 & 0.1 \\\mathrm{f}(\mathrm{x}) & -2.9910 & -2.9999 & -3.0000 & -3.0000 & -2.9999 & -2.9910\end{array} ; \text { limit }=-3.0$

B)
$\begin{array}{r|llllll}\mathrm{x} & -0.1 & -0.01 & -0.001 & 0.001 & 0.01 & 0.1 \\\hline \mathrm{f}(\mathrm{x}) & -1.4970 & -1.4999 & -1.5000 & -1.5000 & -1.4999 & -1.4970\end{array} \text {; limit }=-15.0$

C)
$\begin{array}{c|llllll}\mathrm{x} & -0.1 & -0.01 & -0.001 & 0.001 & 0.01 & 0.1 \\\hline \mathrm{f}(\mathrm{x}) & -1.4970 & -1.4999 & -1.5000 & -1.5000 & -1.4999 & -1.4970\end{array} \text { limit }=\infty$

D)
$\begin{array}{c|llllll}\mathrm{x} & -0.1 & -0.01 & -0.001 & 0.001 & 0.01 & 0.1 \\\hline \mathrm{f}(\mathrm{x}) & -4.9900 & -4.9999 & -5.0000 & -5.0000 & -4.9999 & -4.9900\end{array} \text {; limit }=-5.0$