Question

Complete the tablesIf possible, find the limitState the function value if it existsdescribe as continuous. hole, jump, infinite (unbounded), or oscillating.

Answer 1

`\lim _(x\to6)(x-6)/(x^2-36)`

Complete the table: Substitute the x in the equation for the given value and find the value of y:

`y=(x-6)/(x^2-36)`

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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):

`\begin{gathered} \lim _(x\to6)(x-6)/(x^2-36)=\lim _(x\to6)(x-6)/(x^2-36)\cdot((x+6)/(x+6)) \\ \\ =\lim _(x\to6)(x^2-36)/(x^2-36(x+6))=\lim _(x\to6)(1)/(x+6) \end{gathered}`

Then, evaluate the limit with x=6:

`\lim _(x\to6)(1)/(x+6)=(1)/(6+6)=(1)/(12)`

The limit is 1/12

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