Vertical & Horizontal Asymptotes

Bottom: $x^2 - 4x = x(x - 4) = 0$, so $x = 0$ or $x = 4$.

Top at $x = 0$: $0 + 1 = 1 \neq 0$. Top at $x = 4$: $16 + 1 = 17 \neq 0$. Both are asymptotes.

$x = 0$ and $x = 4$.

Bottom: $x^2 - x - 2 = (x - 2)(x + 1) = 0$, so $x = 2$ or $x = -1$.

Top at $x = -1$: $1 - 4 = -3 \neq 0$, so an asymptote. Top at $x = 2$: $4 - 4 = 0$, both zero, so cancel: $\dfrac{(x - 2)(x + 2)}{(x - 2)(x + 1)} = \dfrac{x + 2}{x + 1}$. The bottom is no longer $0$ at $x = 2$: a hole, not an asymptote.

$x = -1$ only.

Bottom: $x^2 + 4 = 0$ has no real solution ($x^2 + 4 \geq 4$).

No candidates, so nothing to check.

There are none: the bottom is never $0$.

Bottom: $\log x = 0$ at $x = 1$.

Top at $x = 1$: $1 \neq 0$. An asymptote.

$x = 1$.

Top: $6x^2$. Bottom: $2x^2$.

$\dfrac{6x^2}{2x^2} = 3$. Same power, so the limit is $3$.

No $\sqrt{\ }$ or $e^x$, so also $3$.

$y = 3$.

Top: $4x$. Bottom: $x^2$.

$\dfrac{4x}{x^2} = \dfrac{4}{x} \to 0$. The bottom dominates (higher power), so $0$.

Also $0$.

$y = 0$.

Top: $e^{x}$. Bottom: $x^2$. ($e^{x}$ beats any power of $x$.)

$\dfrac{e^{x}}{x^2} \to \infty$: the top dominates, so no horizontal asymptote on the right.

As $x \to -\infty$, $e^{x} \to 0$ while $x^2 + 1 \to \infty$, so $f(x) \to 0$.

$y = 0$, on the left only.

Top: $5x$. Bottom: $\sqrt{x^2} = |x|$.

As $x \to \infty$, $|x| = x$: $\dfrac{5x}{x} = 5$.

As $x \to -\infty$, $|x| = -x$: $\dfrac{5x}{-x} = -5$. Different!

$y = 5$ (on the right) and $y = -5$ (on the left).

$e^{x} - 4 = 0 \iff x = \log 4$.

$\displaystyle\lim_{x \to (\log 4)^+} f(x) = \dfrac{9}{0^+} = +\infty$, so $x = \log 4$ is a vertical asymptote.

Divide by $e^{x}$: $\dfrac{2 + e^{-x}}{1 - 4e^{-x}} \to \dfrac{2}{1} = 2$.

$e^{x} \to 0$: $\dfrac{0 + 1}{0 - 4} = -\dfrac{1}{4}$.

VA: $x = \log 4$. HA: $y = 2$ (right) and $y = -\dfrac{1}{4}$ (left).

$e^{-x} > 0$, so $2 + e^{-x} > 2$ is never $0$: $g$ is defined everywhere.

No candidates, so no vertical asymptote.

$e^{-x} \to 0$: $\dfrac{6}{2 + 0} = 3$.

$e^{-x} \to \infty$: $\dfrac{6}{\text{huge}} \to 0$.

No VA. HA: $y = 3$ (right) and $y = 0$ (left).

The exponent $\dfrac{1}{x}$ is undefined at $x = 0$.

$x \to 0^+$: $\dfrac{1}{x} \to +\infty$, so $e^{1/x} \to \infty$. That one side makes $x = 0$ a VA. (From the left, $\dfrac{1}{x} \to -\infty$ and $e^{1/x} \to 0$.)

$\dfrac{1}{x} \to 0$, so $e^{1/x} \to e^{0} = 1$.

$\dfrac{1}{x} \to 0$ again, so $e^{1/x} \to 1$.

VA: $x = 0$. HA: $y = 1$.

$1 - e^{-x^2} = 0 \iff e^{-x^2} = 1 \iff x = 0$.

For $x \neq 0$, $-x^2 < 0$ so $e^{-x^2} < 1$ and the bottom is $0^+$: $k(x) \to +\infty$ from both sides. VA at $x = 0$.

$-x^2 \to -\infty$, so $e^{-x^2} \to 0$ and $k(x) \to 1$.

$-x^2 \to -\infty$ here too, so $k(x) \to 1$.

VA: $x = 0$. HA: $y = 1$.

$x - 3 = 0$ at $x = 3$.

$9 + 3k - 6 = 0 \Rightarrow k = -1$.

$x^2 - x - 6 = (x - 3)(x + 2)$, so $f(x) = x + 2$ for $x \neq 3$.

$\displaystyle\lim_{x \to 3} (x + 2) = 5$.

$|x + 2| = 0$ at $x = -2$.

$4c - 2 - 6 = 0 \Rightarrow c = 2$.

$2x^2 + x - 6 = (x + 2)(2x - 3)$. For $x > -2$: $f(x) = 2x - 3$. For $x < -2$: $f(x) = -(2x - 3)$.

Right: $2(-2) - 3 = -7$. Left: $-(-7) = 7$. Both finite, so no VA; but $-7 \neq 7$, so the limit does not exist (a jump).

$x - 6 = 0$ at $x = 6$.

$\sqrt{12 + a} - 4 = 0 \Rightarrow 12 + a = 16 \Rightarrow a = 4$.

Conjugate: $\dfrac{(2x + 4) - 16}{(x - 6)(\sqrt{2x + 4} + 4)} = \dfrac{2(x - 6)}{(x - 6)(\sqrt{2x + 4} + 4)} = \dfrac{2}{\sqrt{2x + 4} + 4}$.

$\dfrac{2}{\sqrt{16} + 4} = \dfrac{1}{4}$.