225 lines
7.6 KiB
Plaintext
225 lines
7.6 KiB
Plaintext
#import "./template/template.typ": exercise,notes, theorem, definition, proof, lemma
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#show: notes.with(
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class: "EE1M1",
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lecture: "test",
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number: 2,
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date: "2025-11-25",
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)
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#definition(title: "limit")[
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Let $f$ be a function defined on some open interval that contains the
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number $a$, except possibly at $a$ itself. Then we say that the *limit
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of $f(x)$ as $x$ approaches $L$*, and we write
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$ limits("lim")_(x->a) f(x) = L $
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if for every number $epsilon > 0$ there is a number $delta > 0$ such that
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$ "if" quad 0 < | x-a| < delta quad "then" quad |f(x) - L| < epsilon $
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]
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#definition(title: "left-sided limit")[
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$ lim_(x->a^(-)) f(x) = L $
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if for every number $epsilon > 0$ there is a number $delta > 0$ such that
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$ "if" quad a-delta < x < a quad "then" quad |f(x) - L| < epsilon $
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]
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#definition(title: "right-sided limit")[
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$ lim_(x->a^(+)) f(x) = L $
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if for every number $epsilon > 0$ there is a number $delta > 0$ such that
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$ "if" quad a < x < a+delta quad "then" quad |f(x) - L| < epsilon $
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]
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#definition(title: "infinite limits")[
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Let $f$ be a function defined on some open interval that contains the number
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$a$, except possibly at $a$ itself. Then
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$ limits("lim")_(x->a) f(x) = infinity $
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means that for every positive number $M$ there is a positive number $delta$
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such that
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$ "if" quad 0 < |x-a| < delta quad "then" quad f(x) > M $
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]
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#definition(title: "vertical asymptotes")[
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The vertical line $x = a$ is called a *vertical asymptote* of the curve
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$y=f(x)$ if at least one of the following statements is true:
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#table(
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columns: (33%, 33%, 33%),
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stroke: none,
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[$ lim_(x->a) f(x) = infinity $], [$ lim_(x->a^(-)) f(x) = infinity $],
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[$ lim_(x->a^(+)) f(x) = infinity $], [$ lim_(x->a) f(x) = -infinity $],
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[$ lim_(x->a^(-)) f(x) = -infinity $], [$ lim_(x->a^(+)) f(x) = -infinity $]
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)
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]
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#theorem(title: "Constant Laws")[
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Suppose that $c$ is constant:
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+ $limits("lim")_(x->a) c = c$
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+ $limits("lim")_(x->a) x = a$
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]
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#theorem(title: "Limit Laws")[
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Suppose that $c$ is a constant and the limits:
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$ limits("lim")_(x->a) f(x) quad "and" quad limits("lim")_(x->a) g(x) $
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exist then:
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#block([
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+ *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)
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+ *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)
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+ *constant mult. law*: #v(0pt) $limits("lim")_(x->a) [c f(x)] = c limits("lim")_(x->a) f(x)$ #v(5pt)
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+ *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)
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+ *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)
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+ *power law*: #v(0pt) $limits("lim")_(x->a) [f(x)]^n = [limits("lim")_(x->a)]^n$ where $n$ is a positive integer#v(5pt)
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+ *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)
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if $n$ is even, we assume that $limits("lim")_(x->a) f(x) > 0$
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])
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]
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#proof(title: "limit laws")[\
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*Sum Law*
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Let $limits("lim")_(x->a) f(x) = L$ and $limits("lim")_(x->a) g(x) = M$. then
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$limits("lim")_(x->a) [f(x) + g(x)] = L + M$.
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Let $epsilon > 0$. We must find $delta > 0$ such that
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$ "if" quad 0 < | x - a| < delta quad "then" quad |f(x) + g(x) - (L+M)| < epsilon $
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Using the triangle equality
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$ |(f(x)-L) + (g(x) - M)| <= |f(x) - L| + |g(x) - M| $
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If $|f(x) -L| < epsilon/2$ and $|g(x) - M| < epsilon/2$ then $|f(x) -L + g(x) - M| < epsilon$ by the inequality
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above.
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Since $epsilon/2 > 0$ and $limits("lim")_(x->a) f(x) = L$, there exists a number
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$delta_1 > 0$ such that
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$ "if" quad 0 < | x-a| < delta_1 quad "then" quad |f(x) -L| < epsilon/2 $
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Similarly, with $limits("lim")_(x->a) g(x) = M$, there exists a number $delta_2 > 0$ such that
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$ "if" quad 0 < | x-a| < delta_2 quad "then" quad |f(x) - M| < epsilon/2 $
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Let $delta = min{delta_1, delta_2}$. If $0 < | x - a | < delta$ then
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$0 < |x-a| < delta_1$ and $0 < | x-a| < delta_2$.
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Therefore
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$ |f(x) + g(x) - (L+M)| <= |f(x) -L| + |g(x) - M| < epsilon/2 + epsilon/2 = epsilon $
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]
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#definition(title: "Direct substitution property")[
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If $f$ is a polynomial or a rational function and $a$ is in the domain
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of $f$, then
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$ limits("lim")_(x->a) f(x) = f(a) $
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Functions with this property are called #emph([continuous]) at $a$.
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]
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#theorem(title: "Inequality of limits")[
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If $f(x) <= g(x)$ when $x$ is near $a$ (except possibly at $a$) and the limits
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of $f$ and $g$ both exists as $x$ approaches $a$, then
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$ limits("lim")_(x->a) f(x) <= limits("lim")_(x->a) g(x) $
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]
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#theorem(title: "Squeeze theorem")[
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If $f(x) <= g(x) <= h(x)$ when $x$ is near $a$ (except possibly at $a$) and
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$ limits("lim")_(x->a) f(x) = limits("lim")_(x->a) h(x) = L $
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then
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$ limits("lim")_(x->a) g(x) = L $
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]
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#definition(title: "Limit at infinity")[
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Let $f$ be a function defined on some interval $(a,infinity)$. Then
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$ limits("lim")_(x->infinity) f(x) = L $
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menas that for every $epsilon > 0$ there is a corresponding number $N$ such that
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$ "if" quad x > N quad "then" quad |f(x) - L| < epsilon $
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]
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#exercise(title: "Exercise 9, section 2.2")[
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State the following:
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+ $limits("lim")_(x->-7) f(x) = -infinity$
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+ $limits("lim")_(x->-3) f(x) = infinity$
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+ $limits("lim")_(x->0) f(x) = infinity$
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+ $limits("lim")_(x->6^(-)) f(x) = -infinity $
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+ $limits("lim")_(x->6^(+)) f(x) = infinity $
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]
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#exercise(title: "Exercise 29, section 2.2")[
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Determine the infinite limit:
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$ limits("lim")_(x->5^(+)) (x+1)/(x-5) = infinity $
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]
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#exercise(title: "Exercise 39, section 2.2")[
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Determine the infinite limit:
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$ limits("lim")_(x->0) ln(x^2) - x^(-2) = -infinity $
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]
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#exercise(title: "Exercise 23, section 2.3")[
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Evaluate the limit, if it exists
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$ limits("lim")_(h->0) (sqrt(9+h)-3)/h $
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Using the square root trick:
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$
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limits("lim")_(h->0) (sqrt(9+h)-3)/h dot (sqrt(9+h)+3)/(sqrt(9+h)+3) \
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limits("lim")_(h->0) (9+h - 9)/(h(sqrt(9+h)+3)) = 1/(sqrt(9+h)+3)= 1/6
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$
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]
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#exercise(title: "Exercise 31, section 2.3")[
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Evaluate the limit, if it exists
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$ limits("lim")_(t->0) 1/(t sqrt(1+t)) - 1/t $
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First combinding the terms:
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$ limits("lim")_(t->0) (t- t sqrt(1+t))/(t^2 sqrt(1+t)) = (1-sqrt(1+t))/(t sqrt(1+t)) $
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and then using the square root trick:
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$
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limits("lim")_(t->0) (1-sqrt(1+t))/(t sqrt(1+t)) dot (1+sqrt(1+t))/(1+sqrt(1+t)) = \
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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
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$
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]
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#exercise(title: "Exercise 41, section 2.3")[
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Prove that $limits("lim")_(x->0) x^4 cos(2/x) = 0$.
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Using Squeeze theorem:
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$
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-1 <= cos(2/x) <= 1 \
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-x^4 <= x^4cos(2/x) <= x^4 \
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limits("lim")_(x->0) -x^4 = limits("lim")_(x->0) x^4 = 0
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$
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]
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#exercise(title: "Exercise 47, section 2.3")[
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Find the limit, if it exsists. If the limit does not exist, explain why.
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$ limits("lim")_(x->0^(-)) 1/x - 1/(|x|) = 1/x - (-1/x) = 2/x $
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This limit tends to infinity and does not exist.
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]
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#exercise(title: "Exercise 51, section 2.3")[
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+ $limits("lim")_(x->2^(+)) g(x) = ((x-2)(x+3))/(x-2) = x+3 = 5$
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+ $limits("lim")_(x->2^(-)) g(x) = ((x-2)(x+3))/-(x-2) = -(x+3) = -5$
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+ The limit does not exist, since the one-sided limits diverge.
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]
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#exercise(title: "Exercise 61, section 2.3")[
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$
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limits("lim")_(x->0) (f(x) - 8)/(x-1) = 10 \
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limits("lim")_(x->0) f(x) - 8 = limits("lim")_(x->0) (f(x)-8)/(x-1) dot (x-1) = 10 dot 0 = 0 \
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limits("lim")_(x->0) f(x) - limits("lim")_(x->0) 8 = 0 \
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limits("lim")_(x->0) f(x) = 8
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$
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]
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