Question #317527

Find the limit of the following functions using the tabular method.


  1. lim    x2+x-6x+3/ x+3

π‘₯β†’βˆ’3


  1. lim x2-2x+3/ x+1   

π‘₯β†’βˆ’1






1
Expert's answer
2022-03-27T07:24:56-0400

1.


lim⁑xβ†’βˆ’3x2+xβˆ’6x+3x+3\lim_{x\rightarrow-3} \frac{x^2+x-6x+3}{ x+3}\\

Let f(x)=x2+xβˆ’6x+3x+3f(x) = \frac{x^2+x-6x+3}{ x+3}\\


xβˆ’3.2βˆ’3.19βˆ’3.09βˆ’3.01βˆ’2.99βˆ’2.9βˆ’2.89βˆ’2.8f(x)βˆ’146.2βˆ’153.29βˆ’311.09βˆ’2711.012689.01259.1234.56124.2\def\arraystretch{1.5} \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & -3.2 & -3.19 &-3.09 &-3.01 &-2.99 &-2.9 &-2.89&-2.8 \\ \hline f(x) &-146.2 & -153.29 &-311.09&-2711.01&2689.01&259.1&234.56&124.2\\ \hline \end{array}


The limit of the function as x approaches -1 does not exist, since the value of x from the left approaches smaller negative value and the the value from the right approaches large positive value.


2.


lim⁑xβ†’βˆ’1x2βˆ’2x+3x+1\lim_{x \rightarrow -1} \frac{x^2-2x+3}{x+1}

Let f(x)=x2βˆ’2x+3x+1f(x) = \frac{x^2-2x+3}{x+1}



xβˆ’1.2βˆ’1.11βˆ’1.1βˆ’1.01βˆ’0.99βˆ’0.9βˆ’0.89βˆ’0.8f(x)βˆ’34.2βˆ’58.66βˆ’64.1βˆ’604.01596.0156.150.6526.2\def\arraystretch{1.5} \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & -1.2 & -1.11&-1.1 &-1.01 &-0.99 &-0.9 &-0.89&-0.8 \\ \hline f(x) &-34.2& -58.66 &-64.1&-604.01&596.01&56.1&50.65&26.2\\ \hline \end{array}


The limit of the function as x approaches -1 does not exist, since the value of x from the left approaches smaller negative value and the the value from the right approaches large positive value.


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