# Calculus

Limit Basics

Limits are one of the four main concepts of calculus (Limits, Derivatives, Antiderivatives, and Integrals)
Definition of continuity:

1. $f(c)$ must exist
2. $\lim\limits_{x\to c}{f(x)}$ must exist
3. $\lim\limits_{x\to c}{f(x)}=f(c)$

A limit does not exist if $\lim\limits_{x\to c^+}{f(x)}\neq \lim\limits_{x\to c^-}{f(x)}$; do not confuse this with limits approaching $\pm \infty$
In short, the limit is the value that the function $f(x)$ approaches as $x$ gets arbitrarily closer to a certain value
Horizontal asymptotic behavior for rational functions:
In a rational function (a fraction), if:

a) if the degree of the numerator is larger than the degree of the denominator, then there is no horizontal asymptote
b) if the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator
c) if the degree of the numerator is smaller than the degree of the denominator, then the horizontal asymptote is 0

Horizontal asymptotes are often limits of functions when $x \to \pm \infty$
Misconception: you can and often do cross horizontal asymptotes
Vertical asymptotes can sometimes be limits, other times not
Vertical asymptotes are holy; they can never be crossed

Limit Properties

The essential properties of limits, these can always be used to produce the correct answer. However, sometimes this takes too long. Luckily for you, we have some formulas (that I'll show later on) which can cut down the time per problem.

(I) $\lim\limits_{x\to a}{\left[c\cdot f(x)\right]}= c\cdot \lim\limits_{x\to a}{f(x)}$

(II) $\lim\limits_{x\to a}{\left[f(x) \pm g(x)\right]}= \lim\limits_{x\to a}{f(x)} \pm \lim\limits_{x\to a}{g(x)}$

(III) $\lim\limits_{x\to a}{\left[f(x)\cdot g(x)\right]}=\lim\limits_{x\to a}{f(x)} \cdot \lim\limits_{x\to a}{g(x)}$

(IV) $\lim\limits_{x\to a}{\left[\frac{f(x)}{g(x)}\right]}=\frac{\lim\limits_{x\to a}{f(x)}}{\lim\limits_{x\to a}{g(x)}}; \lim\limits_{x\to a}{g(x)} \neq 0$

(V) $\lim\limits_{x\to a}{\left[f(x)\right]^n}=\left[\lim\limits_{x\to a}{f(x)}\right]^n; n \in \mathbb{R}$

(VI) $\lim\limits_{x\to a}{\left[ \sqrt[n]{f(x)}\right]}=\sqrt[n]{\lim\limits_{x\to a}{f(x)}}$

(VII) $\lim\limits_{x\to a}{c}=c$, where $c$ is a constant

(VIII) $\lim\limits_{x\to a}{x}=a$

(IX)$\lim\limits_{x\to a}{x^n}=a^n$

(X) $\lim\limits_{x\to a} f(g(x))=\lim\limits_{x\to a} f\circ g(x)=f(\lim\limits_{x\to a} g(x))$

Limit Theory

Continuity and differentiability: Differentiability implies continuity, but not vice versa
$\delta - \varepsilon$ definition:

If $\lim\limits_{x \to c}{f(x)}=L$, then there exists a $\delta>\left|x-c\right|$ and an $\varepsilon>\left|f(x)-L\right|$

$\delta$ corresponds to $x$ values while $\varepsilon$ corresponds to $y$ values; $\delta$ is constant while $\varepsilon$ varies

Difference quotients- approximations of derivatives using limits

Forward Difference Quotient: $\lim\limits_{h \to 0}{\frac{f(x+h)-f(x)}{h}}$
Backward Difference Quotient: $\lim\limits_{h \to 0}{\frac{f(x)-f(x-h)}{h}}=\lim\limits_{h \to 0}{\frac{f(x-h)-f(x)}{-h}}$
Symmetric Difference Quotient: $\lim\limits_{h \to 0}{\frac{f(x+h)-f(x-h)}{2h}}$ - most accurate approximation

Squeeze Theorem

The Squeeze Theorem AKA Sandwich Theorem states that if one wanted to take the limit of a function, they could "squeeze" the fuction between two other functions, one that is always greater than and one that is always less than the original. However, at the point of the limit being evaluated, these functions both approach the same limit, which implies that the initial function must also approach this limit. This obviously does not work with limits tending towards $\pm\infty$. More formally, if $g(x)\leq h(x)$ for all $x$ in the neighborhood of $c$, where $x\neq c$, $\displaystyle\lim_{x\to c}g(x)=\displaystyle\lim_{x\to c}h(x)=L$, and $f$ is a function for which $g(x)\leq f(x) \leq h(x)$ for all $x$ in a neighborhood of $c$, then $\displaystyle\lim_{x\to c}f(x)=L$.

The Squeeze Theorem can be implemented algebraically too. Consider $\displaystyle\lim_{x\to0}x^2\sin\left(\frac{1}{x}\right)$. We can evaluate this limit by squeezeing it between two other functions. First, notice that the range of $\sin(x)$ is $[-1,1]$. We can write this as $-1\leq\sin(x)\leq1$. Now we are trying to turn the function in the middle into our original function. Since having $\frac{1}{x}$ does not change the range of the $\sin(x)$ function, we can also write $-1\leq\sin\left(\frac{1}{x}\right)\leq1$. Now all that's left is to multiply the inequality by $x^2$: $-x^2\leq x^2\sin\left(\frac{1}{x}\right)\leq x^2$. Now that we have our original function being squeezed between two other ones, we can evaluate the limit of the entire inequality: $\displaystyle\lim_{x\to0}-x^2\leq\displaystyle\lim_{x\to0}x^2\sin\left(\frac{1}{x}\right)\leq\displaystyle\lim_{x\to0}x^2$. Now we can see that $0\leq\displaystyle\lim_{x\to0}x^2\sin\left(\frac{1}{x}\right)\leq0$, so the answer to our initial problem must be $0$.

Limits at Points

This is largely precalc. All you have to do is substitute in the point in question and evaluate the limit. Sometimes, this will turn into a L'Hospital's limit. It is important to note the continuity of the function when doing this. Some terms to know: monotonic (always increasing or always decreasing on an interval), inverse functions (if $f(x)$ and $g(x)$ are inverses, then $f(g(x))=g(f(x))=x$)

Limits Approaching Infinity

Growth of functions: here is a list of functions from the fastest-growing to the slowest-growing.

$n!$; factorial functions, the fastest-growing
$a^n$; exponential functions
$n^a$; polynomial functions, in order of the power (i.e. $x^3$ grows faster than $x^2$)
$\log (n)$; logarithmic functions, the slowest-growing
$y=a$; constant function, does not grow

End behavior: what limit the function approaches as $x$ tends to infinity
Horizontal asymptote rules often apply: If the degrees of the numerator and denominator are the same, then the limit will be the ratio of leading coefficients, etc.
Tip: you cannot add variable powers (i.e $2^n + 4^n \neq 6^n$)

L'Hospital's Limits

If $\lim\limits_{x \to a}{f(x)} \to \frac{0}{0}, \frac{\infty}{\infty}, \pm \infty \cdot 0, 1^{\infty}, 0^0, \infty^0, \infty-\infty$ (indeterminate forms), and the function that is being evaluated is in a quotient form (i.e. has anumerator and denominator), then L'Hospital's Rule applies. You cannot use L'Hospital's if the function is not made into a quotient
L'Hospital's Rule requires derivatives: derive the numerator and denominator separately (do not use quotient rule), then take the limit again
If limit tends towards an indeterminate form again, then complete however many iterations of L'Hospital's Rule as necessary to obtain an answer
Sometimes, Limits tend to an indeterminate form which are not quotients. In this case, you will need to turn the function into a quotient of two functions by rearranging it or by invoking the tenth limit property stated above, usually with the natural logarithm function in order to bring the exponent down (the limit of a composition is the composition of the limit). For example, if we wanted to evaluate the $\displaystyle\lim_{x\to1^+}{x^{\frac{1}{x-1}}}\to1^\infty$. We can set this limit $=L$ and take the $\ln$ of both sides in order to use L'Hospital's: $\ln L=\ln\left[\displaystyle\lim_{x\to1^+}{x^{\frac{1}{x-1}}}\right]=\displaystyle\lim_{x\to1^+}{\ln x^{\frac{1}{x-1}}}$ $=\displaystyle\lim_{x\to1^+}{\frac{\ln x}{x-1}}\to\frac{0}{0}.$ Now that we have an ideterminate quotient we can apply L'Hospital's rule. Doing so gives us $\displaystyle\lim_{x\to1^+}\frac{\frac{1}{x}}{1}=1$. But since this is equal to the natural logarithm of our limit, we need to exponentiate both sides in order to get the real limit: $L=e^{\ln L}=e^1=e$. So, this limit is equal to $e$.

Limits involving $e$

The number $e$ AKA Euler's constant, can be defined as a limit. Here are some useful formulas to know, all of which can be arrived at by using the limit properties above.

$e=\lim\limits_{x \to \pm \infty}{\left( 1 + \frac{1}{x}\right)^{x}}=\lim\limits_{x \to 0}{(1+x)^{\frac{1}{x}}}$
$e^{a\cdot b}=\lim\limits_{x \to \pm \infty}{\left( 1 + \frac{a}{x}\right)^{bx}}$
$e^{b \cdot c}=\lim\limits_{x \to 0}{(1+bx)^{\frac{c}{x}}}$