How to find vertical asymptotes of a rational function?
1. Factor numerator and denominator. 2. Simplify the function. 3. Find values where the denominator is zero and the numerator is non-zero.
How to show $x=a$ is a vertical asymptote using limits?
1. Evaluate $\lim_{x\to a^+} f(x)$. 2. Evaluate $\lim_{x\to a^-} f(x)$. 3. Show that at least one of these limits is $\pm \infty$.
How to determine if a discontinuity is removable or a vertical asymptote?
1. Factor and simplify the function. 2. If a factor cancels, it's a removable discontinuity. 3. If a factor remains in the denominator, it's a vertical asymptote.
How to find the vertical asymptotes of $f(x) = \frac{x+2}{x^2 - 4}$?
1. Factor: $f(x) = \frac{x+2}{(x+2)(x-2)}$. 2. Simplify: $f(x) = \frac{1}{x-2}$. 3. VA at $x=2$.
Steps to find vertical asymptotes of $f(x) = \tan(x)$?
1. Rewrite as $f(x) = \frac{\sin(x)}{\cos(x)}$. 2. Find where $\cos(x) = 0$. 3. $x = \frac{(2n+1)\pi}{2}$, where n is an integer.
How to find the vertical asymptotes of $f(x) = \frac{x^2 - 1}{x-1}$?
1. Factor: $f(x) = \frac{(x-1)(x+1)}{x-1}$. 2. Simplify: $f(x) = x+1$. 3. Removable discontinuity at x=1, no vertical asymptote.
How to find the limit of a rational function as x approaches a vertical asymptote?
1. Determine the sign of the function as x approaches from the left and right. 2. The limit will be either positive or negative infinity.
How to solve for vertical asymptotes of $f(x) = \frac{1}{x^2 + 1}$?
1. Set denominator equal to zero: $x^2 + 1 = 0$. 2. Solve for x: $x^2 = -1$. 3. No real solutions, so no vertical asymptotes.
How to find vertical asymptotes of $f(x) = \frac{e^x}{x}$?
1. Check where the denominator is zero: $x=0$. 2. Verify the numerator is not zero at $x=0$: $e^0 = 1 \neq 0$. 3. Vertical asymptote at $x=0$.
How to determine the behavior of $f(x) = \frac{1}{x-2}$ as x approaches 2?
1. Check limit from the right: $\lim_{x\to 2^+} \frac{1}{x-2} = \infty$. 2. Check limit from the left: $\lim_{x\to 2^-} \frac{1}{x-2} = -\infty$. 3. Vertical asymptote at $x=2$.
What are the differences between removable and non-removable discontinuities?
Removable: can be 'fixed' by redefining the function. Non-removable: cannot be fixed; often vertical asymptotes.
Compare and contrast $\lim_{x\to a^+} f(x) = \infty$ and $\lim_{x\to a^-} f(x) = \infty$.
Both indicate the function approaches infinity, but the first approaches from the right, the second from the left.
Compare and contrast $\lim_{x\to a^+} f(x) = \infty$ and $\lim_{x\to a^-} f(x) = -\infty$.
The first approaches positive infinity from the right, the second approaches negative infinity from the left, indicating a vertical asymptote.
What is the difference between a limit existing and a vertical asymptote existing?
A limit exists if the function approaches a finite value. A vertical asymptote exists if the function approaches infinity.
Compare the behavior of a function near a vertical asymptote to its behavior near a hole (removable discontinuity).
Near a vertical asymptote, the function approaches infinity. Near a hole, the function approaches a finite value, but is undefined at that point.
Compare the graphs of $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x^2}$ near x=0.
f(x) has different signs on either side of x=0. g(x) is always positive.
Compare the domains of $f(x) = \ln(x)$ and $g(x) = \frac{1}{x}$.
f(x) is defined for x > 0. g(x) is defined for all x except x = 0.
Contrast the behavior of polynomial functions with rational functions in terms of vertical asymptotes.
Polynomial functions do not have vertical asymptotes, while rational functions can have vertical asymptotes where the denominator is zero.
Compare the limits of $f(x) = \frac{1}{x-a}$ and $g(x) = x-a$ as x approaches a.
The limit of f(x) is infinite, indicating a vertical asymptote. The limit of g(x) is 0.
Contrast the behavior of a function at a vertical asymptote where the power of the factor in the denominator is even versus odd.
Even power: function approaches same infinity from both sides. Odd power: function approaches opposite infinities from each side.
If $x=a$ is a vertical asymptote, what is the limit?
$\lim_{x\to a} f(x) = \pm\infty$, $\lim_{x\to a^+} f(x) = \pm\infty$, or $\lim_{x\to a^-} f(x) = \pm\infty$
What is the limit definition related to vertical asymptotes?
If $\lim_{x\to a^+} f(x) = \pm \infty$ or $\lim_{x\to a^-} f(x) = \pm \infty$, then $x=a$ is a vertical asymptote.
What is $\lim_{x\to 0^+} ln(x)$?
$\lim_{x\to 0^+} ln(x) = -\infty$
What is the general form of a function with a vertical asymptote at x=a?
f(x) = $\frac{g(x)}{x-a}$, where g(a) != 0
If $f(x) = \frac{1}{x+c}$, what is the vertical asymptote?
x = -c
If $f(x) = \frac{1}{(x-a)^n}$ where n is even, what is $\lim_{x \to a} f(x)$?
$\infty$
If $f(x) = \frac{1}{(x-a)^n}$ where n is odd, what are $\lim_{x \to a^+} f(x)$ and $\lim_{x \to a^-} f(x)$?
$\lim_{x \to a^+} f(x) = \infty$ and $\lim_{x \to a^-} f(x) = -\infty$
What is the limit of $\frac{1}{x}$ as $x$ approaches 0 from the right?
$\lim_{x\to 0^+} \frac{1}{x} = \infty$
What is the limit of $\frac{1}{x}$ as $x$ approaches 0 from the left?
$\lim_{x\to 0^-} \frac{1}{x} = -\infty$
If $f(x) = \frac{p(x)}{q(x)}$ has a vertical asymptote at $x=a$, what must be true of $p(a)$ and $q(a)$?
$q(a) = 0$ and $p(a) \neq 0$
