Consider limits where \(x\) approaches \(+\infty\) or \(-\infty\).
Special cases:
Theorem. For all constant numbers \( c \), \[ \lim_{x \to \infty} c = c \quad \text{and} \quad \lim_{x \to -\infty} c = c. \]
Example. \[ \lim_{x \to \infty} 23 = 23 \quad \text{and} \quad \lim_{x \to -\infty} 100 = 100. \]
Theorem. For all \( n > 0 \), \[ \lim_{x \to \infty} x^n = \infty \quad \text{and} \quad \lim_{x \to \infty} x^{-n} = \lim_{x \to \infty} \frac{1}{x^n} = 0 \]
Example. \[ \lim_{x \to \infty} x^3 = \infty \quad \text{and} \quad \lim_{x \to \infty} x^{-3} = \lim_{x\to \infty}\frac{1}{x^3} = 0 \]
Theorem. If \( n \) is a positive whole number, \[ \lim_{x \to -\infty} x^n = \begin{cases} \infty & \text{if } n \text{ is even} \\ -\infty & \text{if } n \text{ is odd} \end{cases} \quad \text{and} \quad \lim_{x \to -\infty} x^{-n} = \lim_{x \to -\infty} \frac{1}{x^n} = 0 \]
Example. \[ \lim_{x \to -\infty} x^3 = -\infty \quad \text{and} \quad \lim_{x \to -\infty} x^4 = \infty \]
Example. \[ \lim_{x \to -\infty} x^{-3} = \lim_{x\to -\infty}\frac{1}{x^3} = 0, \quad \text{and} \quad \lim_{x \to -\infty} x^{-4} = \lim_{x\to -\infty}\frac{1}{x^4} = 0 \]
Here the rational functions are of the form polynomial over polynomial.
Example. \[ \frac{3x^2 + 2x + 1}{x^2 + 1}, \quad \frac{x^5 + 2x + 1}{x^7 + x^4 + x^2 + 3}, \quad \frac {2x^2 + 1}{x + 1} \]
Theorem. The asymptotic behavior of a rational function depends only on the leading terms of its numerator and denominator. If \( a_n, b_m \neq 0 \), then \[ \lim_{x \to \infty} \frac{a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0}{b_m x^m + b_{m-1} x^{m-1} + \cdots + b_0} = \frac{a_n}{b_m}\cdot(\lim_{x \to \infty} x^{n-m}). \] \[ \lim_{x \to -\infty} \frac{a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0}{b_m x^m + b_{m-1} x^{m-1} + \cdots + b_0} = \frac{a_n}{b_m}\cdot(\lim_{x \to -\infty} x^{n-m}). \]
How do we know that a continuous function \(f(x) = 0\) has a root?
One way is to use the Intermediate Value Theorem.
The Intermediate Value Theorem (IVT) says that a continuous function cannot skip values.
In other words, if a continuous function \(f\) takes on values \(f(a)\) and \(f(b)\), then it takes on all values between \(f(a)\) and \(f(b)\).
Theorem. ]
Then there exists a number \( c \) in \((a, b)\) such that \( f(c) = L \).
How does the Intermediate Value Theorem help us to see there is a root in the interval?
Theorem. If \( f \) is continuous on \([a, b]\) and \( f(a) \cdot f(b) < 0 \), then there exists a number \( c \) in \((a, b)\) such that \( f(c) = 0 \).
`Once we know that a continuous function has a root, how can we find the root more accurately?
One way is to use the Bisection Method.
Here is the process:
Question. When can we stop the process?
Answer. When the interval is small enough. The Problem will provide the length of the intervals.
Definition.The difference quotient of a function \( f \) at a number \( x\) is \[ \frac{f(a) - f(x)}{a - x}, \] where \(x\) is a number from \(a\).
Hence, the difference quotient is the slope of the secant line through the points \((a, f(a))\) and \((x, f(x))\).
Now, if we denote \(h = a - x\), then \(a = x + h\). The difference quotient becomes \[ \frac{f(a) - f(x)}{a - x} = \frac{f(x+h) - f(x)}{h}. \] And we will have an alternative form of the difference quotient.
Definition. The difference quotient of a function \( f \) at a number \( x \) is \[ \frac{f(x+h) - f(x)}{h}. \]
Definition. The derivative of a function \( f \) at a number \( x \) is \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \] \[ f'(x) = \lim_{a \to x} \frac{f(a) - f(x)}{a - x}. \]
Remark. The derivative of a function at a number is the slope of the tangent line to the graph of the function at that number.
\(f'(x)\) is also denoted by \(\frac{df}{dx}\), \(\frac{dy}{dx}\), or \(\frac{d}{dx} f(x)\).
Definition. The derivative of a function \( f \) at a number \( x \) is \[ \frac{d}{dx}f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \] \[ \frac{d}{dx}f(x) = \lim_{a \to x} \frac{f(a) - f(x)}{a - x}. \]
Theorem. If \( f(x) = mx + b \) is a linear function, then \( f'(a) = m \) for all \( a \) in its domain.
If \( f(x) = b \) is a constant function, then \( f'(a) = 0 \) for all \( a \) in its domain.
Example. If \( f(x) = \frac23x + 3 \) is a linear function define on \(x\in [5, 10]\), then \( f'(a) = \frac23 \) for all \( a \in [5, 10]\).
Example. If \( f(x) = 100 \) is a constant function, whose domain is \((-\infty, \infty)\). Then ,\( f'(a) = 0 \) for all \( a \).
Theorem. For all exponents \( n \): \[ \frac{d}{dx} x^n = n x^{n-1} \]
Example. \[ \frac{d}{dx} x^3 = 3x^2 \]
Example. \[ \frac{d}{dx} x^{-1} = -1x^{-2} \]
Example. \[ \frac{d}{dx} x^{-\frac{1}{2}} = -\frac12x^{-\frac32} \]
Theorem. Assume that \( f \) and \( g \) are differentiable. Then \( f + g \) and \( f - g \) are differentiable, and \[ (f + g)' = f' + g', \quad (f - g)' = f' - g' \]
Example. \[ \frac{d}{dx} (x^2 + 3x) = \frac{d}{dx} x^2 + \frac{d}{dx} 3x = 2x + 3. \]
Theorem. For any constant \( c \), \( c\cdot f \) is differentiable, and \[ (c\cdot f)' = c\cdot f' \]
Example. \[ \frac{d}{dx} 3x^2 = 3\frac{d}{dx} x^2 = 3\cdot 2x = 6x. \]
Theorem. For all \( x \), \[ \frac{d}{dx} e^x = e^x \]
Theorem. If \( f \) is differentiable at \( x = c \), then \( f \) is continuous at \( x = c \).
Example. If \( f \) is not continuous at \( x = c \), then \( f \) is not differentiable at \( x = c \).
Theorem. If \( f \) and \( g \) are differentiable, then \( f\cdot g \) is differentiable, and \[ (f\cdot g)' = f'\cdot g + f\cdot g' \]
Example. \[ \frac{d}{dx} (x^2\cdot e^x) = \frac{d}{dx} x^2\cdot e^x + x^2\cdot \frac{d}{dx} e^x = 2x\cdot e^x + x^2\cdot e^x \]
Theorem. If \( f \) and \( g \) are differentiable, then \( \frac{f}{g} \) is differentiable, and \[ \left(\frac{f}{g}\right)' = \frac{f'\cdot g - f\cdot g'}{g^2} \]
Example. \[ \frac{d}{dx} \frac{x^2}{e^x} = \frac{2x\cdot e^x - x^2\cdot e^x}{(e^x)^2} \]