NEED KO PO ANSWER SALAMAT

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NEED KO PO ANSWER SALAMAT​

 

Appoximate this limit using table of values. Show your complete solution

\\begin{array}{*{20}{c}}{}&{Limit}&{from}&{left}&{Limit}&{from}&{right}\\x&{0.9}&{0.99}&{0.999}&{1.1}&{1.01}&{1.001}\\{f(x)}&{ - 15.263}&{ - 150.251}&{ - 1500.250}&{14.762}&{149.751}&{1499.75}\end{array}\

\\mathop {\lim }\limits_{x \to 1} \frac{{x + 2}}{{{x^2} - 1}} = \frac{{1 + 2}}{{{1^2} - 1}} = \frac{3}{0}\

= undefined

Limit from left

\\begin{array}{l}x = 0.9\\\frac{{x + 2}}{{{x^2} - 1}} = \frac{{0.9 + 2}}{{{{(0.9)}^2} - 1}} = \frac{{2.9}}{{0.81 - 1}} = \frac{{2.9}}{{ - 0.19}} = - 15.263\end{array}\

\\begin{array}{l}x = 0.99\\\frac{{x + 2}}{{{x^2} - 1}} = \frac{{0.99 + 2}}{{{{(0.99)}^2} - 1}} = \frac{{2.99}}{{0.9801 - 1}} = \frac{{2.99}}{{ - 0.0199}} = - 150.251\end{array}\

\\begin{array}{l}x = 0.999\\\frac{{x + 2}}{{{x^2} - 1}} = \frac{{0.999 + 2}}{{{{(0.999)}^2} - 1}} = \frac{{2.999}}{{0.998001 - 1}} = \frac{{2.999}}{{ - 0.002}} = - 1500.250\end{array}\

Limit from right

\\begin{array}{l}x = 1.1\\\frac{{x + 2}}{{{x^2} - 1}} = \frac{{1.1 + 2}}{{{{(1.1)}^2} - 1}} = \frac{{3.1}}{{1.21 - 1}} = \frac{{3.1}}{{0.21}} = 14.762\end{array}\

\\begin{array}{l}x = 1.01\\\frac{{x + 2}}{{{x^2} - 1}} = \frac{{1.01 + 2}}{{{{(1.01)}^2} - 1}} = \frac{{3.01}}{{1.0201 - 1}} = \frac{{3.01}}{{0.201}} = 149.751\end{array}\

\\begin{array}{l}x = 1.001\\\frac{{x + 2}}{{{x^2} - 1}} = \frac{{1.001 + 2}}{{{{(1.001)}^2} - 1}} = \frac{{3.001}}{{1.002001 - 1}} = \frac{{3.001}}{{0.002001}} = 1499.75\end{array}\

Use the limit laws to evaluate the following

1.)

\\mathop {\lim }\limits_{x \to 3} (5x - 2) = ((5)(3) - 2 = 15 - 2 = 13\

2.)

\\mathop {\lim }\limits_{x \to 2} (7x + 8) = ((7)(2) + 8 = 14 + 8 = 22\

3.)

\\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} + 3x - 2}}{{{x^3} + 1}} = \frac{{3{{(2)}^2} + 3(2) - 2}}{{{{(2)}^3} + 1}} = \frac{{12 + 6 - 2}}{{8 + 1}} = \frac{{16}}{9}\

4.)

\\mathop {\lim }\limits_{x \to 1} \frac{{(4x + 3)}}{{(2x - 1)}} = \frac{{(4)(1) + 3)}}{{(2(1) - 1)}} = \frac{7}{1} = 7\

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