From f1550639c0855f49b4b6af87d838b4547bcb95f9 Mon Sep 17 00:00:00 2001 From: Folkert Kevelam Date: Sun, 23 Nov 2025 22:45:50 +0100 Subject: [PATCH] Add parts of the second lecture --- new_notes/Calculus_Lecture_2.typ | 224 +++++++++++++++++++++++++++++++ 1 file changed, 224 insertions(+) create mode 100644 new_notes/Calculus_Lecture_2.typ diff --git a/new_notes/Calculus_Lecture_2.typ b/new_notes/Calculus_Lecture_2.typ new file mode 100644 index 0000000..692f9c7 --- /dev/null +++ b/new_notes/Calculus_Lecture_2.typ @@ -0,0 +1,224 @@ +#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 + $ +]