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Use the Table of Values of F to Estimate the Limit

Question 181

Multiple Choice

Use the table of values of f to estimate the limit.
-  Let f(x) =x24x5x27x+10, find limx5f(x) \text { Let } f(x) =\frac{x^{2}-4 x-5}{x^{2}-7 x+10} \text {, find } \lim _{x \rightarrow 5} f(x) \text {. }
x4.94.994.9995.0015.015.1f(x) \begin{array}{c|l|l|l|l|l|l}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & & & & & &\end{array}


A)
x4.94.994.9995.0015.015.1f(x) 2.13452.10332.10032.09972.09672.0677 limit =2.1\begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 2.1345 & 2.1033 & 2.1003 & 2.0997 & 2.0967 & 2.0677\end{array} \text { limit }=2.1

B)
x4.94.994.9995.0015.015.1f(x) 1.93451.90331.90031.89971.89671.8677 limit = 1.9\begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 1.9345 & 1.9033 & 1.9003 & 1.8997 & 1.8967 & 1.8677\end{array} \text { limit = } 1.9

C)
x4.94.994.9995.0015.015.1f(x) 2.03452.00332.00031.99971.99671.9677 limit = 2 \begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 2.0345 & 2.0033 & 2.0003 & 1.9997 & 1.9967 & 1.9677\end{array} \text { limit = 2 }

D)
x4.94.994.9995.0015.015.1f(x) 0.57750.57200.57150.57140.57080.5652 limit =0.5714\begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 0.5775 & 0.5720 & 0.5715 & 0.5714 & 0.5708 & 0.5652\end{array} \text { limit }=0.5714

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