Given the function below
[tex]\lim _{x\rightarrow-2}(\frac{x+2}{x^3+8})[/tex]
To find the limit function above, let try to simplify the denominator as shown below using sum of cube expansion
[tex]x^3+8=x^3+3^3[/tex]
Please note that
[tex]\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}[/tex]
Substituting the expansion of the sum of cubes into the limit gives
[tex]\begin{gathered} \lim _{x\rightarrow-2}(\frac{x+2}{x^3+8})=\lim _{x\rightarrow-2}(\frac{x+2}{x^3+3^3}) \\ \lim _{x\rightarrow-2}(\frac{x+2}{x^3+8})=\lim _{x\rightarrow-2}(\frac{x+2}{(x+2)(x^2-2x+4)}) \\ \lim _{x\rightarrow-2}(\frac{x+2}{x^3+8})=\lim _{x\rightarrow-2}(\frac{1}{x^2-2x+4}) \end{gathered}[/tex]
Let us put the limiting value of -2 as shown below
[tex]\begin{gathered} \lim _{x\rightarrow-2}(\frac{1}{x^2-2x+4}),\text{put x= -2} \\ =\frac{1}{(-2)^2-2(-2)+4} \\ =\frac{1}{4+4+4} \\ =\frac{1}{12} \end{gathered}[/tex]
Hence, the limit of the given limiting function is 1/12, OPTION B
