Calculus I - Limits
27 flashcards · Mathematics · by luckydog789 (harrison holbrook)
Sum law of limits
limx→a(f(x)±g(x))=limx→af(x)±limx→ag(x)\lim_{x\to a}(f(x) \pm g(x))=\lim_{x\to a}f(x) \pm \lim_{x\to a}g(x)limx→a(f(x)±g(x))=limx→af(x)±limx→ag(x)
Constant multiple law of limits
limx→a(cf(x))=climx→af(x) \lim_{x \to a} (cf(x)) = c\lim_{x \to a} f(x)limx→a(cf(x))=climx→af(x)
Product law of limits
limx→a(f(x)g(x)))=(limx→af(x))(limx→ag(x)) \lim_{x \to a} (f(x)g(x))) = (\lim_{x \to a} f(x))(\lim_{x \to a} g(x))limx→a(f(x)g(x)))=(limx→af(x))(limx→ag(x))
Quotient law of limits
limx→af(x)g(x)=limx→af(x)limx→ag(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}limx→ag(x)f(x)=limx→ag(x)limx→af(x) if limx→ag(x)≠0\text{if } \lim_{x \to a}g(x)\neq 0if limx→ag(x)=0
Power law of limits
limx→af(x)n=(limx→af(x))n \lim_{x \to a} f(x)^{n} = (\lim_{x \to a} f(x))^{n}limx→af(x)n=(limx→af(x))n
Root law of limits
limx→af(x)n=limx→af(x)n \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)}limx→anf(x) =nlimx→af(x) limx→af(x)1n=(limx→af(x))1n \lim_{x \to a} f(x)^{\frac{1}{n}} = {(\lim_{x \to a} f(x))}^{\frac{1}{n}}limx→af(x)n1=(limx→af(x))n1 0 \text{ for all x near a}"> if n is even, f(x)>0 for all x near a\text{if n is even, } f(x) > 0 \text{ for all x near a}if n is even, f(x)>0 for all x near a
If p(x)p(x)p(x) and q(x)q(x)q(x) are polynomials, and q(a)≠0q(a) \neq 0q(a)=0
The following flashcards use these statements
Polynomial limit law
limx→ap(x)=p(a)\lim_{x \to a} p(x)=p(a)limx→ap(x)=p(a)
Rational function limit law
limx→ap(x)q(x)=p(a)q(a)\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}limx→aq(x)p(x)=q(a)p(a)
The squeeze theorem
Assume the functions f, g, and h satisfy the inequality f(x)≤g(x)≤h(x)f(x)\leq g(x) \leq h(x)f(x)≤g(x)≤h(x) for all x near a (except possibly at a). If limx→af(x)=limx→ah(x)=L\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = Llimx→af(x)=limx→ah(x)=L , then limx→ag(x)=L\lim_{x \to a} g(x)=Llimx→ag(x)=L
Rational function limit law when limx→ap(x)≠0\lim_{x\to a}p(x)\neq 0limx→ap(x)=0 and limx→aq(x)=0\lim_{x\to a}q(x)=0limx→aq(x)=0
limx→ap(x)q(x)=±∞\lim_{x\to a}\frac{p(x)}{q(x)}=\pm \infty limx→aq(x)p(x)=±∞ ,depending on the sign of limx→ap(x)\lim_{x\to a}p(x)limx→ap(x)
Assume limx→af(x) \lim_{x \to a} f(x)limx→af(x) and limx→ag(x) \lim_{x \to a} g(x)limx→ag(x) exist. The following are true, when c∈Rc \in \mathbb{R}c∈R , and 0"> n∈Z∧n>0n \in \mathbb{Z} \wedge n > 0n∈Z∧n>0
The following flashcards use these statements
One sided limit equality law
if and only if limx→a+f(x)=limx→a−f(x)\lim_{x\to a^{+}}f(x) = \lim_{x\to a^{-}}f(x)limx→a+f(x)=limx→a−f(x) exists, then limx→af(x)\lim_{x\to a}f(x)limx→af(x) exists
End behavior of functions as x→∞x\to\inftyx→∞ , 0"> n∈Z∧n>0n \in \mathbb{Z} \wedge n > 0n∈Z∧n>0
The following flashcards use these statements
f(x)f(x)f(x) gets arbitrarily large
limx→∞f(x)=∞\lim_{x\to \infty}{f(x)}=\inftylimx→∞f(x)=∞
f(x)f(x)f(x) gets arbitrarily close to some number L
limx→∞f(x)=L\lim_{x\to \infty}{f(x)}=Llimx→∞f(x)=L
f(x)f(x)f(x) resembles 1xn\frac{1}{x^n}xn1 ( 0"> n∈Z∧n>0n \in \mathbb{Z} \wedge n > 0n∈Z∧n>0 )
limx→∞f(x)=0\lim_{x\to\infty}f(x)=0limx→∞f(x)=0
If the degree of polynomial p(x)p(x)p(x) is less than the degree of polynomial q(x)q(x)q(x)
limx→∞p(x)q(x)=0\lim_{x\to\infty}\frac{p(x)}{q(x)}=0limx→∞q(x)p(x)=0
If the degree of polynomial p(x)p(x)p(x) is greater than the degree of polynomial q(x)q(x)q(x)
limx→∞p(x)q(x)=±∞\lim_{x\to\infty}\frac{p(x)}{q(x)}=\pm\inftylimx→∞q(x)p(x)=±∞
Suppose p(x)=amxm...p(x)=a_mx^m...p(x)=amxm... , q(x)=bnxn...q(x)=b_nx^n...q(x)=bnxn... , and that the degree of both polynomials are the same. What is the limit of p(x)q(x)\frac{p(x)}{q(x)}q(x)p(x) as x→∞x\to\inftyx→∞
limx→∞p(x)q(x)=ambn\lim_{x\to\infty}{\frac{p(x)}{q(x)}}=\frac{a_m}{b_n}limx→∞q(x)p(x)=bnam
Limit of a periodic function
The exact limit does not exist, but the limit can be seen as ranging from the maxima and minima of the periodic function
End behavior of exe^xex
limx→∞ex=∞\lim_{x\to\infty}e^x=\inftylimx→∞ex=∞ limx→−∞ex=0\lim_{x\to -\infty}e^x=0limx→−∞ex=0
End behavior of e−xe^{-x}e−x
limx→∞e−x=0\lim_{x\to\infty}{e^{-x}}=0limx→∞e−x=0 limx→−∞e−x=∞\lim_{x\to -\infty}{e^{-x}}=\inftylimx→−∞e−x=∞
End behavior of ln(x)\ln(x)ln(x)
limx→∞ln(x)=∞\lim_{x\to\infty}\ln(x)=\inftylimx→∞ln(x)=∞ limx→0+ln(x)=−∞\lim_{x\to 0^+}\ln(x)=-\inftylimx→0+ln(x)=−∞
End behavior of sin(x)x\frac{\sin(x)}{x}xsin(x)
Even though sin(x)\sin(x)sin(x) has no limit, since it is bounded between -1 and 1, it can be treated as both limx→±∞sin(x)x=0\lim_{x\to \pm \infty}\frac{\sin(x)}{x}=0limx→±∞xsin(x)=0
End behavior of cos(x)−1x\frac{\cos(x)-1}{x}xcos(x)−1
Even though cos(x)\cos(x)cos(x) has no limit, since it is bounded between -1 and 1, it can be treated as both limx→0cos(x)−1x=0\lim_{x\to 0}\frac{\cos(x)-1}{x}=0limx→0xcos(x)−1=0
The number eee as a limit
limh→0(1+h)1h=e\lim_{h \to 0}(1+h)^{\frac{1}{h}}=elimh→0(1+h)h1=e limn→∞(1+1n)n=e\lim_{n \to \infty}(1+ \frac{1}{n})^n=elimn→∞(1+n1)n=e