Question

A function is said to have a vertical asymptote wherever the limit on the left or right (or both) is either positive or negative infinity.

For example, the function f(x)= \frac{-3(x+2)}{x^2+4x+4} has a vertical asymptote at x=-2. For each of the following limits, enter either 'P' for positive infinity, 'N' for negative infinity, or 'D' when the limit simply does not exist.
\displaystyle{ \lim_{x\to -2^-} \frac{-3(x+2)}{x^2+4x+4} = }
\displaystyle{ \lim_{x\to -2^+} \frac{-3(x+2)}{x^2+4x+4} =}
\displaystyle{ \lim_{x\to -2} \frac{-3(x+2)}{x^2+4x+4} =}

Answer 1

`\lim_(x\to -2^-) (-3(x+2))/(x^2+4x+4) \\ \\ \text{Let} \\ \\x=-2-h\ where\ h\to 0 \\ \\ \text{so} \\ \\\lim_(h\to 0) (-3(-2-h+2))/((-2-h)^2+4(-2-h)+4) =\lim_(h \to 0)(-3* -h)/(4+h^2+4h-8-4h+4) \\ \\ \text{We get} \\ \\\lim_(h\to 0)(3h)/(h^2)=+\infty\Rightarrow P`



`\lim_(x\to -2^+) (-3(x+2))/(x^2+4x+4) \\ \\ \text{Let} \\ \\x=-2+h\ where\ h\to 0 \\ \\ \text{so} \\ \\\lim_(h\to 0) (-3(-2+h+2))/((-2+h)^2+4(-2+h)+4) =\lim_(h \to 0)(-3* h)/(4+h^2-4h-8+4h+4) \\ \\ \text{We get} \\ \\\lim_(h\to 0)(-3h)/(h^2)=-\infty\Rightarrow P`


`\lim_(x\to -2) (-3(x+2))/(x^2+4x+4) =(-3(-2+2))/((-2)^2+4(-2)+4) = (-3*0)/(0) = (-0)/(0) \Rightarrow D`