Question

Calculus early transcendental functions. Need help to see the work to solve

Answer 1

Given the function below

`\lim _(x\rightarrow-2)((x+2)/(x^3+8))`

To find the limit function above, let try to simplify the denominator as shown below using sum of cube expansion

`x^3+8=x^3+3^3`

Please note that

`\begin{gathered} x^3+3^3=(x+2)(x^2-2x+2^2) \\ x^3+3^3=(x+2)(x^2-2x+4) \end{gathered}`

Substituting the expansion of the sum of cubes into the limit gives

`\begin{gathered} \lim _(x\rightarrow-2)((x+2)/(x^3+8))=\lim _(x\rightarrow-2)((x+2)/(x^3+3^3)) \\ \lim _(x\rightarrow-2)((x+2)/(x^3+8))=\lim _(x\rightarrow-2)((x+2)/((x+2)(x^2-2x+4))) \\ \lim _(x\rightarrow-2)((x+2)/(x^3+8))=\lim _(x\rightarrow-2)((1)/(x^2-2x+4)) \end{gathered}`

Let us put the limiting value of -2 as shown below

`\begin{gathered} \lim _(x\rightarrow-2)((1)/(x^2-2x+4)),\text{put x= -2} \\ =(1)/((-2)^2-2(-2)+4) \\ =(1)/(4+4+4) \\ =(1)/(12) \end{gathered}`

Hence, the limit of the given limiting function is 1/12, OPTION B