Complete the table: Substitute the x in the equation for the given value and find the value of y:
[tex]y=\frac{x-6}{x^2-36}[/tex][tex]\begin{gathered} y=\frac{5.9-6}{(5.9)^2-36}=\frac{-0.1}{34.81-36}=\frac{-0.1}{-1.19}=0.08403 \\ \\ y=\frac{5.99-6}{(5.99)^2-36}=\frac{-0.01}{35.8801-36}=\frac{-0.01}{-0.1199}=0.083402 \\ \\ y=\frac{5.999-6}{(5.999)^2-36}=\frac{-0.001}{35.988001-36}=\frac{-0.001}{-0.011999}=0.0833402 \\ \\ y=\frac{6-6}{(6)^2-36}=\frac{0}{36-36}=\frac{0}{0}=\text{undefined} \\ \\ y=\frac{6.001-6}{(6.001)^2-36}=\frac{0.001}{36.012001-36}=\frac{0.001}{0.012001}=0.083326 \\ \\ y=\frac{6.01-6}{(6.01)^2-36}=\frac{0.01}{36.1201-36}=\frac{0.01}{0.1201}=0.083263 \\ \\ y=\frac{6.1-6}{(6.1)^2-36}=\frac{0.1}{37.21-36}=\frac{0.1}{1.21}=0.082644 \end{gathered}[/tex]-------------------------
With the data of the table you can see that the limit when x tends to 6 is 0.083 (or 1/12). You can get an accurate answer to the limit as follow:
To find the limit multiply numerator and denominator by the conjugate of the numerator (x+6):
[tex]\begin{gathered} \lim _{x\to6}\frac{x-6}{x^2-36}=\lim _{x\to6}\frac{x-6}{x^2-36}\cdot(\frac{x+6}{x+6}) \\ \\ =\lim _{x\to6}\frac{x^2-36}{x^2-36(x+6)}=\lim _{x\to6}\frac{1}{x+6} \end{gathered}[/tex]Then, evaluate the limit with x=6:
[tex]\lim _{x\to6}\frac{1}{x+6}=\frac{1}{6+6}=\frac{1}{12}[/tex]The limit is 1/12
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