Question

1. determine the vertical asymptotes of each of the following functions. then calculate the one sided limits of the functions at each asymptote.

i) ( f(x) = \frac{x - 1}{(x + 2)^2} )

ii) ( g(x) = \frac{x^2 - x - 6}{x^2 - 2x - 8} )

iii) ( h(x) = \frac{-x}{(x + 4)(x - 3)} )

Answer:

Part (i): Function \( f(x) = \frac{x - 1}{(x + 2)^2} \)

Step 1: Find Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is non - zero at those points).
Set the denominator equal to zero: \((x + 2)^2=0\). Solving for \(x\), we get \(x=-2\).
Check the numerator at \(x = - 2\): When \(x=-2\), the numerator \(x - 1=-2 - 1=-3
eq0\). So, \(x=-2\) is a vertical asymptote.

Step 2: Calculate One - Sided Limits at \(x=-2\)

• **Left - hand limit (\(x

ightarrow - 2^{-}\))**:
As \(x
ightarrow - 2^{-}\), \(x+2
ightarrow0^{-}\) (since \(x\) is slightly less than \(-2\)), so \((x + 2)^2
ightarrow0^{+}\) (squaring a negative number close to 0 gives a positive number close to 0). The numerator \(x - 1
ightarrow-2 - 1=-3\) (a non - zero constant).
So, \(\lim_{x
ightarrow - 2^{-}}\frac{x - 1}{(x + 2)^2}=\frac{-3}{0^{+}}=-\infty\) (wait, no: \(\frac{\text{negative}}{\text{positive small}}=-\infty\)? Wait, numerator is \(-3\) (negative), denominator is \((x + 2)^2\), when \(x
ightarrow - 2^{-}\), \(x+2\) is negative and close to 0, squaring it makes it positive and close to 0. So \(\frac{-3}{\text{positive small}}\) is \(-\infty\)? Wait, no: \(-3\) divided by a very small positive number is a very large negative number. But wait, let's re - evaluate.
Wait, \(x
ightarrow - 2^{-}\): let \(x=-2 - h\) where \(h
ightarrow0^{+}\). Then \(x + 2=-h\), \((x + 2)^2=h^{2}\), and \(x-1=-2 - h-1=-3 - h\). So \(\frac{x - 1}{(x + 2)^2}=\frac{-3 - h}{h^{2}}\). As \(h
ightarrow0^{+}\), the numerator \(
ightarrow - 3\) and the denominator \(
ightarrow0^{+}\). So \(\lim_{h
ightarrow0^{+}}\frac{-3 - h}{h^{2}}=-\infty\) (since the numerator is negative and the denominator is positive and approaches 0).

• **Right - hand limit (\(x

ightarrow - 2^{+}\))**:
Let \(x=-2 + h\) where \(h
ightarrow0^{+}\). Then \(x + 2=h\), \((x + 2)^2=h^{2}\), and \(x - 1=-2+h - 1=-3 + h\). So \(\frac{x - 1}{(x + 2)^2}=\frac{-3 + h}{h^{2}}\). As \(h
ightarrow0^{+}\), the numerator \(
ightarrow - 3\) and the denominator \(
ightarrow0^{+}\). So \(\lim_{h
ightarrow0^{+}}\frac{-3 + h}{h^{2}}=-\infty\) (because the numerator is negative and the denominator is positive and approaches 0). Wait, actually, when \(x
ightarrow - 2^{+}\), \(x + 2
ightarrow0^{+}\), so \((x + 2)^2
ightarrow0^{+}\), and \(x-1
ightarrow - 3\). So \(\frac{-3}{0^{+}}=-\infty\)? Wait, no, \(-3\) divided by a very small positive number is a very large negative number. But wait, let's check with a test value. Let \(x=-1.9\) (close to \(-2\) from the right). Then \((x + 2)=0.1\), \((x + 2)^2 = 0.01\), \(x-1=-2.9\). Then \(\frac{-2.9}{0.01}=-290\), which is negative and large in magnitude. As \(x\) gets closer to \(-2\) from the right, the value gets more negative. So \(\lim_{x
ightarrow - 2^{+}}\frac{x - 1}{(x + 2)^2}=-\infty\) and \(\lim_{x
ightarrow - 2^{-}}\frac{x - 1}{(x + 2)^2}=-\infty\)

Part (ii): Function \( g(x)=\frac{x^{2}-x - 6}{x^{2}-2x - 8} \)

Step 1: Simplify the Function (if possible)

Factor the numerator and the denominator:

• Numerator: \(x^{2}-x - 6=(x - 3)(x + 2)\)
• Denominator: \(x^{2}-2x - 8=(x - 4)(x + 2)\)

So, \(g(x)=\frac{(x - 3)(x + 2)}{(x - 4)(x + 2)}\), for \(x
eq - 2\). We can cancel out the common factor \((x + 2)\) (since \(x
eq - 2\)) and get \(g(x)=\frac{x - 3}{x - 4}\), \(x
eq - 2\).

Step 2: Find Vertical Asymptotes

Vertical asymptotes occur where the denominator of the simplified function is zero (and the numerator is non - zero at those points).
Set the denominator of the simplified function equal to zero: \(x - 4=0\), so \(x = 4\).
Check the numerator at \(x = 4\): When \(x = 4\), the numerator \(x-3=4 - 3 = 1
eq0\). So, \(x = 4\) is a vertical asymptote. The point \(x=-2\) is a hole (a removable discontinuity), not a vertical asymptote.

Step 3: Calculate One - Sided Limits at \(x = 4\)

• **Left - hand limit (\(x

ightarrow4^{-}\))**:
As \(x
ightarrow4^{-}\), \(x - 4
ightarrow0^{-}\) (since \(x\) is slightly less than 4) and \(x-3
ightarrow4 - 3=1\). So, \(\lim_{x
ightarrow4^{-}}\frac{x - 3}{x - 4}=\frac{1}{0^{-}}=-\infty\) (because a positive number divided by a negative number close to 0 is a large negative number).

• **Right - hand limit (\(x

ightarrow4^{+}\))**:
As \(x
ightarrow4^{+}\), \(x - 4
ightarrow0^{+}\) (since \(x\) is slightly more than 4) and \(x-3
ightarrow4 - 3 = 1\). So, \(\lim_{x
ightarrow4^{+}}\frac{x - 3}{x - 4}=\frac{1}{0^{+}}=\infty\) (because a positive number divided by a positive number close to 0 is a large positive number).

Part (iii): Function \( h(x)=\frac{-x}{(x + 4)(x - 3)} \)

Step 1: Find Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is non - zero at those points).
Set the denominator equal to zero: \((x + 4)(x - 3)=0\). Solving for \(x\), we get \(x=-4\) or \(x = 3\).

• Check the numerator at \(x=-4\): When \(x=-4\), the numerator \(-x = 4

eq0\).

• Check the numerator at \(x = 3\): When \(x = 3\), the numerator \(-x=-3

eq0\). So, \(x=-4\) and \(x = 3\) are vertical asymptotes.

Step 2: Calculate One - Sided Limits at \(x=-4\)

• **Left - hand limit (\(x

ightarrow - 4^{-}\))**:
As \(x
ightarrow - 4^{-}\), \(x + 4
ightarrow0^{-}\) (since \(x\) is slightly less than \(-4\)) and \(x-3
ightarrow-4 - 3=-7\). The numerator \(-x
ightarrow4\). So, \(\frac{-x}{(x + 4)(x - 3)}=\frac{4}{(0^{-})(-7)}=\frac{4}{0^{+}}\) (because \((0^{-})\times(-7)=0^{+}\)). So \(\lim_{x
ightarrow - 4^{-}}\frac{-x}{(x + 4)(x - 3)}=\infty\).

• **Right - hand limit (\(x

ightarrow - 4^{+}\))**:
As \(x
ightarrow - 4^{+}\), \(x + 4
ightarrow0^{+}\) (since \(x\) is slightly more than \(-4\)) and \(x-3
ightarrow-4 - 3=-7\). The numerator \(-x
ightarrow4\). So, \(\frac{-x}{(x + 4)(x - 3)}=\frac{4}{(0^{+})(-7)}=\frac{4}{0^{-}}\). So \(\lim_{x
ightarrow - 4^{+}}\frac{-x}{(x + 4)(x - 3)}=-\infty\).

Step 3: Calculate One - Sided Limits at \(x = 3\)

• **Left - hand limit (\(x

ightarrow3^{-}\))**:
As \(x
ightarrow3^{-}\), \(x - 3
ightarrow0^{-}\) (since \(x\) is slightly less than 3) and \(x + 4
ightarrow3 + 4 = 7\). The numerator \(-x
ightarrow-3\). So, \(\frac{-x}{(x + 4)(x - 3)}=\frac{-3}{(7)(0^{-})}=\frac{-3}{0^{-}}=\infty\) (because a negative number divided by a negative number close to 0 is a large positive number).

• **Right - hand limit (\(x

ightarrow3^{+}\))**:
As \(x
ightarrow3^{+}\), \(x - 3
ightarrow0^{+}\) (since \(x\) is slightly more than 3) and \(x + 4
ightarrow3 + 4 = 7\). The numerator \(-x
ightarrow-3\). So, \(\frac{-x}{(x + 4)(x - 3)}=\frac{-3}{(7)(0^{+})}=\frac{-3}{0^{+}}=-\infty\) (because a negative number divided by a positive number close to 0 is a large negative number).

Summary of Results

(i) \( f(x)=\frac{x - 1}{(x + 2)^2} \)

• Vertical Asymptote: \(x=-2\)
• \(\lim_{x

ightarrow - 2^{-}}f(x)=-\infty\), \(\lim_{x
ightarrow - 2^{+}}f(x)=-\infty\)

(ii) \( g(x)=\frac{x^{2}-x - 6}{x^{2}-2x - 8} \)

• Vertical Asymptote: \(x = 4\) ( \(x=-2\) is a hole)
• \(\lim_{x

ightarrow4^{-}}g(x)=-\infty\), \(\lim_{x
ightarrow4^{+}}g(x)=\infty\)

(iii) \( h(x)=\frac{-x}{(x + 4)(x - 3)} \)

• Vertical Asymptotes: \(x=-4\) and \(x = 3\)
• At \(x=-4\): \(\lim_{x

ightarrow - 4^{-}}h(x)=\infty\), \(\lim_{x
ightarrow - 4^{+}}h(x)=-\infty\)

• At \(x = 3\): \(\lim_{x

ightarrow3^{-}}h(x)=\infty\), \(\lim_{x
ightarrow3^{+}}h(x)=-\infty\)