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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} =}

Respuesta :

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


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

[tex]\lim_{x\to -2} \frac{-3(x+2)}{x^2+4x+4} =\frac{-3(-2+2)}{(-2)^2+4(-2)+4} = \frac{-3\times0}{0} = \frac{-0}{0} \Rightarrow D[/tex]

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