Use the table of values of f to estimate the limit.
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\text { Let } f(x)=\frac{x-2}{x^{2}-6 x+8} \text {, find } \lim _{x \rightarrow 2} f(x)

\begin{array}{c|l|l|l|l|l|l}\mathrm{x} & 1.9 & 1.99 & 1.999 & 2.001 & 2.01 & 2.1 \\\hline \mathrm{f}(\mathrm{x}) & & & & & &\end{array}

\begin{array}{c|cccccc}\mathrm{x} & 1.9 & 1.99 & 1.999 & 2.001 & 2.01 & 2.1 \\\hline \mathrm{f}(\mathrm{x}) & -0.3762 & -0.3975 & -0.3998 & -0.4003 & -0.4025 & -0.4263\end{array} \text {; limit }=-0.4

\begin{array}{c|cccccc}\mathrm{x} & 1.9 & 1.99 & 1.999 & 2.001 & 2.01 & 2.1 \\\hline \mathrm{f}(\mathrm{x}) & 0.4762 & 0.4975 & 0.4998 & 0.5003 & 0.5025 & 0.5263\end{array} \text {; limit }=0.5

\begin{array}{c|cccccc}\mathrm{x} & 1.9 & 1.99 & 1.999 & 2.001 & 2.01 & 2.1 \\\hline \mathrm{f}(\mathrm{x}) & -0.4762 & -0.4975 & -0.4998 & -0.5003 & -0.5025 & -0.5263\end{array} \text { limit }=-0.5

\begin{array}{c|cccccc}\mathrm{x} & 1.9 & 1.99 & 1.999 & 2.001 & 2.01 & 2.1 \\\hline \mathrm{f}(\mathrm{x}) & -0.5762 & -0.5975 & -0.5998 & -0.6003 & -0.6025 & -0.6263\end{array} ; \text { limit }=-0.6

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Not Answered