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vesna_86 [32]
4 years ago
9

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} =}
Mathematics
1 answer:
lubasha [3.4K]4 years ago
7 0
\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


\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

\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

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Altitudes $\overline{ax}$ and $\overline{by}$ of acute triangle $abc$ intersect at $h$. if $\angle ahb = 132^\circ$, then what i
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-------------------------------------------------------

See the attached image for a visual of the problem and answer.

Based on the diagram, we have angle AHB = 132 degrees (given) equal in measure to angle XHY since these two angles are vertical angles. 

Angles HYC and HXC are right angles due to the nature of AX and BY being altitudes. Recall that altitudes are segments that go from one vertex to the opposite side and they are perpendicular to the opposite side.

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The angle we want to find is angle ACB, which is the same as angle YCX. This angle is the missing angle of the quadrilateral HXCY.

For any quadrilateral, the four angles must add to 360 degrees. 

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