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 rational functions (that is functions that are fractions of polynomials) this occurs whenever the numerator is non - zero and the denominator is 0. for example, the function ( f(x)=\frac{-3(x + 1)}{x^2 + 4x + 4} ) has a vertical asymptote at ( x = - 2 ) since at ( x = - 2 ) the denominator is 0. find each of the following limits. ( lim_{x
ightarrow - 2}\frac{-3(x + 1)}{x^2 + 4x + 4}=square ) help (limits) ( lim_{x
ightarrow - 2^+}\frac{-3(x + 1)}{x^2 + 4x + 4}=square ) help (limits) ( lim_{x
ightarrow - 2}\frac{-3(x + 1)}{x^2 + 4x + 4}=square ) help (limits)

Explanation:

First, we simplify the denominator. The denominator \(x^2 + 4x + 4\) can be factored as a perfect square trinomial.

Step 1: Factor the denominator

The denominator \(x^2 + 4x + 4\) factors to \((x + 2)^2\). So the function becomes \(f(x)=\frac{-3(x + 1)}{(x + 2)^2}\).

Step 2: Analyze the limit as \(x\to - 2\) (two - sided limit)

We check the behavior of the numerator and the denominator as \(x\to - 2\).
- The numerator: When \(x=-2\), the numerator is \(-3(-2 + 1)=-3\times(-1) = 3 eq0\).
- The denominator: \((x + 2)^2=( - 2+2)^2 = 0\), and \((x + 2)^2>0\) for \(x eq - 2\) (since it's a square of a real number, the square of a non - zero real number is positive).

As \(x\to - 2\), the numerator approaches \(3\) (a non - zero constant) and the denominator approaches \(0\) from the positive side (because \((x + 2)^2>0\) for \(x eq - 2\)). So \(\lim_{x\to - 2}\frac{-3(x + 1)}{x^2+4x + 4}=\lim_{x\to - 2}\frac{-3(x + 1)}{(x + 2)^2}\). Since the numerator approaches \(3\) and the denominator approaches \(0\) from the positive side, the limit is \(+\infty\)? Wait, no. Wait, the numerator is \(-3(x + 1)\). Let's re - evaluate the numerator as \(x\to - 2\):

When \(x\to - 2\), \(x+1\to - 2 + 1=-1\), so \(-3(x + 1)\to - 3\times(-1)=3\). The denominator \((x + 2)^2\to0\) and \((x + 2)^2>0\) for \(x eq - 2\). So \(\frac{-3(x + 1)}{(x + 2)^2}\to\frac{3}{0^{+}}\), which is \(+\infty\)? Wait, no, wait the numerator is \(-3(x + 1)\). Wait, when \(x\) is near \(-2\), say \(x=-2+\epsilon\) where \(\epsilon\) is a small non - zero number.

For the left - hand limit (\(x\to - 2^{-}\)): Let \(x=-2-\epsilon\) where \(\epsilon>0\) and \(\epsilon\to0\). Then \(x + 2=-\epsilon\), so \((x + 2)^2=\epsilon^{2}>0\). The numerator: \(-3(x + 1)=-3(-2-\epsilon + 1)=-3(-1-\epsilon)=3 + 3\epsilon\to3\) as \(\epsilon\to0\). So \(\frac{-3(x + 1)}{(x + 2)^2}=\frac{3 + 3\epsilon}{\epsilon^{2}}\to+\infty\) as \(\epsilon\to0^{+}\) (since both numerator and denominator are positive and denominator approaches \(0\)).

For the right - hand limit (\(x\to - 2^{+}\)): Let \(x=-2+\epsilon\) where \(\epsilon>0\) and \(\epsilon\to0\). Then \(x + 2=\epsilon\), so \((x + 2)^2=\epsilon^{2}>0\). The numerator: \(-3(x + 1)=-3(-2+\epsilon + 1)=-3(-1+\epsilon)=3-3\epsilon\to3\) as \(\epsilon\to0\). So \(\frac{-3(x + 1)}{(x + 2)^2}=\frac{3-3\epsilon}{\epsilon^{2}}\to+\infty\) as \(\epsilon\to0^{+}\) (since numerator approaches \(3\) (positive) and denominator approaches \(0\) (positive)).

And for the two - sided limit \(\lim_{x\to - 2}\frac{-3(x + 1)}{x^2+4x + 4}\), since both left - hand and right - hand limits are \(+\infty\), the two - sided limit is also \(+\infty\). Wait, but let's check again. Wait, the numerator: when \(x=-2\), \(-3(-2 + 1)=3\), denominator is \(0\), and denominator is positive for \(x eq - 2\). So the function approaches \(+\infty\) as \(x\) approaches \(-2\) from both sides.

Wait, maybe I made a mistake in the sign. Wait, the numerator is \(-3(x + 1)\). Let's take \(x=-1.9\) (right of \(-2\)): \(x + 1=-0.9\), \(-3(x + 1)=2.7\), denominator \((-1.9 + 2)^2=(0.1)^2 = 0.01\), so \(2.7/0.01 = 270\) (positive). Take \(x=-2.1\) (left of \(-2\)): \(x + 1=-1.1\), \(-3(x + 1)=3.3\), denominator \((-2.1+2)^2=(-0.1)^2 = 0.01\), so \(3.3/0.01 = 330\) (positive). So both left and right limits are \(+\infty\), and the two - sided limit is \(+\infty\).

For \(\lim_{x\to - 2}\frac{-3(x + 1)}{x^2+4x + 4}\)

Step 1: Factor denominator

\(x^2 + 4x + 4=(x + 2)^2\), so the function is \(\frac{-3(x + 1)}{(x + 2)^2}\)

Step 2: Analyze numerator and denominator at \(x=-2\)

- Numerator at \(x = - 2\): \(-3(-2 + 1)=3 eq0\)
- Denominator at \(x=-2\): \((-2 + 2)^2 = 0\), and \((x + 2)^2>0\) for \(x eq - 2\)

Since numerator approaches \(3\) (non - zero) and denominator approaches \(0\) from positive side, \(\lim_{x\to - 2}\frac{-3(x + 1)}{x^2+4x + 4}=+\infty\)

For \(\lim_{x\to - 2^{-}}\frac{-3(x + 1)}{x^2+4x + 4}\)

Step 1: Substitute \(x=-2-\epsilon\) (\(\epsilon>0,\epsilon\to0\))

\(x + 2=-\epsilon\), so \((x + 2)^2=\epsilon^{2}\)
\(x + 1=-2-\epsilon+1=-1-\epsilon\), so \(-3(x + 1)=3 + 3\epsilon\)

Step 2: Find the limit

\(\lim_{\epsilon\to0^{+}}\frac{3 + 3\epsilon}{\epsilon^{2}}\). As \(\epsilon\to0^{+}\), numerator \(\to3\) (positive), denominator \(\to0^{+}\), so the limit is \(+\infty\)

For \(\lim_{x\to - 2^{+}}\frac{-3(x + 1)}{x^2+4x + 4}\)

Step 1: Substitute \(x=-2+\epsilon\) (\(\epsilon>0,\epsilon\to0\))

\(x + 2=\epsilon\), so \((x + 2)^2=\epsilon^{2}\)
\(x + 1=-2+\epsilon+1=-1+\epsilon\), so \(-3(x + 1)=3-3\epsilon\)

Step 2: Find the limit

\(\lim_{\epsilon\to0^{+}}\frac{3-3\epsilon}{\epsilon^{2}}\). As \(\epsilon\to0^{+}\), numerator \(\to3\) (positive), denominator \(\to0^{+}\), so the limit is \(+\infty\)

Answer:

\(\lim_{x\to - 2}\frac{-3(x + 1)}{x^2+4x + 4}=\boldsymbol{+\infty}\)

\(\lim_{x\to - 2^{-}}\frac{-3(x + 1)}{x^2+4x + 4}=\boldsymbol{+\infty}\)

\(\lim_{x\to - 2^{+}}\frac{-3(x + 1)}{x^2+4x + 4}=\boldsymbol{+\infty}\)