#import "./template/template.typ": exercise,notes, theorem, definition, proof, lemma

#show: notes.with(
  class: "EE1M1",
  lecture: "test",
  number: 2,
  date: "2025-11-25",
)

#definition(title: "limit")[
  Let $f$ be a function defined on some open interval that contains the
  number $a$, except possibly at $a$ itself. Then we say that the *limit
  of $f(x)$ as $x$ approaches $L$*, and we write

  $ limits("lim")_(x->a) f(x) = L $

  if for every number $epsilon > 0$ there is a number $delta > 0$ such that

  $ "if" quad 0 < | x-a| < delta quad "then" quad |f(x) - L| < epsilon $
]

#definition(title: "left-sided limit")[
  $ lim_(x->a^(-)) f(x) = L $
  if for every number $epsilon > 0$ there is a number $delta > 0$ such that
  $ "if" quad a-delta < x < a quad "then" quad |f(x) - L| < epsilon $
]

#definition(title: "right-sided limit")[
  $ lim_(x->a^(+)) f(x) = L $
  if for every number $epsilon > 0$ there is a number $delta > 0$ such that
  $ "if" quad a < x < a+delta quad "then" quad |f(x) - L| < epsilon $
]

#definition(title: "infinite limits")[
  Let $f$ be a function defined on some open interval that contains the number
  $a$, except possibly at $a$ itself. Then
  $ limits("lim")_(x->a) f(x) = infinity $

  means that for every positive number $M$ there is a positive number $delta$
  such that

  $ "if" quad 0 < |x-a| < delta quad "then" quad f(x) > M $
]

#definition(title: "vertical asymptotes")[
  The vertical line $x = a$ is called a *vertical asymptote* of the curve
  $y=f(x)$ if at least one of the following statements is true:

  #table(
    columns: (33%, 33%, 33%),
    stroke: none,
    [$ lim_(x->a) f(x) = infinity $], [$ lim_(x->a^(-)) f(x) = infinity $],
    [$ lim_(x->a^(+)) f(x) = infinity $], [$ lim_(x->a) f(x) = -infinity $],
    [$ lim_(x->a^(-)) f(x) = -infinity $], [$ lim_(x->a^(+)) f(x) = -infinity $]
  )
]

#theorem(title: "Constant Laws")[
  Suppose that $c$ is constant:

  + $limits("lim")_(x->a) c = c$
  + $limits("lim")_(x->a) x = a$
]

#theorem(title: "Limit Laws")[
  Suppose that $c$ is a constant and the limits:

  $ limits("lim")_(x->a) f(x) quad "and" quad limits("lim")_(x->a) g(x) $
  exist then:

  #block([
  + *sum law*: #v(0pt) $limits("lim")_(x->a) [f(x) + g(x)] = limits("lim")_(x->a) f(x) + limits("lim")_(x->a) g(x)$ #v(5pt)
  + *difference law*: #v(0pt) $limits("lim")_(x->a) [f(x) - g(x)] = limits("lim")_(x->a) f(x) - limits("lim")_(x->a) g(x)$ #v(5pt)
  + *constant mult. law*: #v(0pt) $limits("lim")_(x->a) [c f(x)] = c limits("lim")_(x->a) f(x)$ #v(5pt)
  + *product law*: #v(0pt) $limits("lim")_(x->a) [f(x)g(x)] = limits("lim")_(x->a) f(x) dot limits("lim")_(x->a) g(x)$ #v(5pt)
  + *quotient law*: #v(0pt) $limits("lim")_(x->a) f(x)/g(x) = (limits("lim")_(x->a) f(x))/(limits("lim")_(x->a) g(x)) quad "if" quad limits("lim")_(x->a) g(x) eq.not 0$ #v(5pt)
  + *power law*: #v(0pt) $limits("lim")_(x->a) [f(x)]^n = [limits("lim")_(x->a)]^n$ where $n$ is a positive integer#v(5pt)
  + *root law*: #v(0pt) $limits("lim")_(x->a) root(n, f(x)) = root(n, limits("lim")_(x->a) f(x))$ where $n$ is a positive integer #v(0pt)
    if $n$ is even, we assume that $limits("lim")_(x->a) f(x) > 0$
  ])
]

#proof(title: "limit laws")[\
  *Sum Law*

  Let $limits("lim")_(x->a) f(x) = L$ and $limits("lim")_(x->a) g(x) = M$. then
  $limits("lim")_(x->a) [f(x) + g(x)] = L + M$. 

  Let $epsilon > 0$. We must find $delta > 0$ such that
  $ "if" quad 0 < | x - a| < delta  quad "then" quad |f(x) + g(x) - (L+M)| < epsilon $

  Using the triangle equality

  $ |(f(x)-L) + (g(x) - M)| <= |f(x) - L| + |g(x) - M| $

  If $|f(x) -L| < epsilon/2$ and $|g(x) - M| < epsilon/2$ then $|f(x) -L + g(x) - M| < epsilon$ by the inequality
  above.

  Since $epsilon/2 > 0$ and $limits("lim")_(x->a) f(x) = L$, there exists a number
  $delta_1 > 0$ such that

  $ "if" quad 0 < | x-a| < delta_1 quad "then" quad |f(x) -L| < epsilon/2 $

  Similarly, with $limits("lim")_(x->a) g(x) = M$, there exists a number $delta_2 > 0$ such that

  $ "if" quad 0 < | x-a| < delta_2 quad "then" quad |f(x) - M| < epsilon/2 $

  Let $delta = min{delta_1, delta_2}$. If $0 < | x - a | < delta$ then
  $0 < |x-a| < delta_1$ and $0 < | x-a| < delta_2$.

  Therefore

  $ |f(x) + g(x) - (L+M)| <= |f(x) -L| + |g(x) - M| < epsilon/2 + epsilon/2 = epsilon $
]

#definition(title: "Direct substitution property")[
  If $f$ is a polynomial or a rational function and $a$ is in the domain
  of $f$, then

  $ limits("lim")_(x->a) f(x) = f(a) $

  Functions with this property are called #emph([continuous]) at $a$.
]

#theorem(title: "Inequality of limits")[
  If $f(x) <= g(x)$ when $x$ is near $a$ (except possibly at $a$) and the limits
  of $f$ and $g$ both exists as $x$ approaches $a$, then

  $ limits("lim")_(x->a) f(x) <= limits("lim")_(x->a) g(x) $
]

#theorem(title: "Squeeze theorem")[
  If $f(x) <= g(x) <= h(x)$ when $x$ is near $a$ (except possibly at $a$) and
  $ limits("lim")_(x->a) f(x) = limits("lim")_(x->a) h(x) = L $
  then
  $ limits("lim")_(x->a) g(x) = L $
]

#definition(title: "Limit at infinity")[
  Let $f$ be a function defined on some interval $(a,infinity)$. Then
  $ limits("lim")_(x->infinity) f(x) = L $

  menas that for every $epsilon > 0$ there is a corresponding number $N$ such that

  $ "if" quad x > N quad "then" quad |f(x) - L| < epsilon $
]

#exercise(title: "Exercise 9, section 2.2")[
  State the following:
  + $limits("lim")_(x->-7) f(x) = -infinity$
  + $limits("lim")_(x->-3) f(x) = infinity$
  + $limits("lim")_(x->0) f(x) = infinity$
  + $limits("lim")_(x->6^(-)) f(x) = -infinity $
  + $limits("lim")_(x->6^(+)) f(x) = infinity $
]

#exercise(title: "Exercise 29, section 2.2")[
  Determine the infinite limit:
  $ limits("lim")_(x->5^(+)) (x+1)/(x-5) = infinity $
]

#exercise(title: "Exercise 39, section 2.2")[
  Determine the infinite limit:
  $ limits("lim")_(x->0) ln(x^2) - x^(-2) = -infinity $
]

#exercise(title: "Exercise 23, section 2.3")[
  Evaluate the limit, if it exists
  $ limits("lim")_(h->0) (sqrt(9+h)-3)/h $
  Using the square root trick:

  $
    limits("lim")_(h->0) (sqrt(9+h)-3)/h dot (sqrt(9+h)+3)/(sqrt(9+h)+3) \
    limits("lim")_(h->0) (9+h - 9)/(h(sqrt(9+h)+3)) = 1/(sqrt(9+h)+3)= 1/6
  $
]

#exercise(title: "Exercise 31, section 2.3")[
  Evaluate the limit, if it exists
  $ limits("lim")_(t->0) 1/(t sqrt(1+t)) - 1/t $
  First combinding the terms:
  $ limits("lim")_(t->0) (t- t sqrt(1+t))/(t^2 sqrt(1+t)) = (1-sqrt(1+t))/(t sqrt(1+t)) $
  and then using the square root trick:

  $
    limits("lim")_(t->0) (1-sqrt(1+t))/(t sqrt(1+t)) dot (1+sqrt(1+t))/(1+sqrt(1+t)) = \
    limits("lim")_(t->0) (1-(1+t))/(t sqrt(1+t)(1+sqrt(1+t))) = -1/(sqrt(1+t)(1+sqrt(1+t))) = -1/2
  $
]

#exercise(title: "Exercise 41, section 2.3")[
  Prove that $limits("lim")_(x->0) x^4 cos(2/x) = 0$.

  Using Squeeze theorem:

  $
    -1 <= cos(2/x) <= 1 \
    -x^4 <= x^4cos(2/x) <= x^4 \
    limits("lim")_(x->0) -x^4 = limits("lim")_(x->0) x^4 = 0
  $
]

#exercise(title: "Exercise 47, section 2.3")[
  Find the limit, if it exsists. If the limit does not exist, explain why.

  $ limits("lim")_(x->0^(-)) 1/x - 1/(|x|) = 1/x - (-1/x) = 2/x $

  This limit tends to infinity and does not exist.
]

#exercise(title: "Exercise 51, section 2.3")[
  + $limits("lim")_(x->2^(+)) g(x) = ((x-2)(x+3))/(x-2) = x+3 = 5$
  + $limits("lim")_(x->2^(-)) g(x) = ((x-2)(x+3))/-(x-2) = -(x+3) = -5$
  + The limit does not exist, since the one-sided limits diverge.
]

#exercise(title: "Exercise 61, section 2.3")[
  $
    limits("lim")_(x->0) (f(x) - 8)/(x-1) = 10 \
    limits("lim")_(x->0) f(x) - 8 = limits("lim")_(x->0) (f(x)-8)/(x-1) dot (x-1) = 10 dot 0 = 0 \
    limits("lim")_(x->0) f(x) - limits("lim")_(x->0) 8 = 0 \
    limits("lim")_(x->0) f(x) = 8
  $
]