Mehlala ea Lipotso tse Buisanang ka Meeli ea Mesebetsi ea Algebraic
Moeli oa mosebetsi oa algebra ke mohopolo oa motheo ho calculus, ho hlahloba boitšoaro ba mosebetsi ha boleng ba oona bo fetohang bo atamela ntlha e itseng. Ho utloisisa meeli ho bohlokoa lits'ebetsong tse fapaneng tsa lipalo, ho kenyeletsoa tlhahlobo ea lipalo le ho etsa mohlala. Sengoloa sena se tla hlalosa mohopolo oa moeli oa mosebetsi oa algebra ka ho fana ka mehlala e 'maloa ea mathata le litharollo tsa ona.
Khopolo ea Motheo ea Meeli ea Mesebetsi ea Algebraic
Pele re kena mathateng a mohlala, ha re hlahlobeng mohopolo oa motheo oa meeli. Moeli oa mosebetsi \( f(x) \) ha \( x \) o atamela boleng \( a \) o bontšoa ke:
\[ \lim_{x \ho isa ho a} f(x) = L \]
e leng se bolelang hore boleng ba \( f(x) \) bo atamela \( L \) jwalo ka \( x \) bo atamela \( a \).
Lipotso tsa Mehlala le Puisano
Mohlala oa Potso ea 1: Moeli oa Mesebetsi e Bonolo ea Algebraic
Fumana boleng bo latelang ba moedi:
\[ \lim_{x \ho isa ho 2} (3x + 4) \]
Puisano:
Bakeng sa mosebetsi o otlolohileng o kang ona, re ka nkela boleng ba \( x \) sebaka ka ho toba ka 2:
\[ \lim_{x \ho isa ho 2} (3x + 4) = 3(2) + 4 = 6 + 4 = 10 \]
Kahoo, \( \lim_{x \to 2} (3x + 4) = 10 \).
Mohlala Potso ea 2: Moeli oa Mosebetsi oa Polynomial
Fumana boleng bo latelang ba moedi:
\[ \lim_{x \ho isa ho -1} (x^2 + 2x + 1) \]
Puisano:
Jwalo ka potsong ya pele, re ka nkela boleng ba \( x \) sebaka ka -1 ka ho toba mosebetsing wa polynomial:
\[ \lim_{x \ho isa ho -1} (x^2 + 2x + 1) = (-1)^2 + 2(-1) + 1 \]
\[ = 1 – 2 + 1 \]
\[ = 0 \]
Kahoo, \( \lim_{x \to -1} (x^2 + 2x + 1) = 0 \).
Mohlala Potso ea 3: Moeli oa Mesebetsi ea Algebraic ka Likaroloana
Fumana boleng bo latelang ba moedi:
\[ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} \]
Puisano:
Haeba re kenya \( x = 3 \) ka ho toba mosebetsing, re fumana foromo e sa tsejweng \( \frac{0}{0} \). Ho rarolla sena, re hloka ho etsa di-factorize:
\[ \frac{x^2 – 9}{x – 3} = \frac{(x – 3)(x + 3)}{x – 3} \]
Pele o hlakola \( x – 3 \), hlokomela hore \( x \neq 3 \), e le hore re ka hlakola \( x – 3 \):
\[ = x + 3 \]
Jwale nka sebaka sa \( x = 3 \):
\[ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 3 + 3 = 6 \]
Kahoo, \( \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 6 \).
Mohlala Bothata ba 4: Meeli ea Mesebetsi ka Metso
Fumana boleng bo latelang ba moedi:
\[ \lim_{x \ho isa ho 4} \sqrt{2x + 1} \]
Puisano:
Kaha mosebetsi o metsong ke mosebetsi o tswelang pele, re ka nkela boleng ba \( x = 4 \ sebaka ka ho toba):
\[ \lim_{x \ho isa ho 4} \sqrt{2x + 1} = \sqrt{2(4) + 1} \]
\[ = \sqrt{8 + 1} \]
\[ = \sqrt{9} \]
\[ = 3 \]
Kahoo, \( \lim_{x \to 4} \sqrt{2x + 1} = 3 \).
Mohlala Potso ea 5: Moeli oa Mesebetsi ea Algebraic ka ho Lokisa Maikutlo
Fumana boleng bo latelang ba moedi:
\[ \lim_{x \to 1} \frac{\sqrt{x + 3} – 2}{x – 1} \]
Puisano:
Phetolo e tobileng \( x = 1 \) e tla hlahisa sebopeho se sa hlakang \( \frac{0}{0} \). Kahoo re hloka ho fana ka mabaka. Atisa nomoro le denominator ka dipara tsa tsona tse tsamaellanang:
\[ \frac{\sqrt{x + 3} – 2}{x – 1} \times \frac{\sqrt{x + 3} + 2}{\sqrt{x + 3} + 2} = \frac{(\sqrt{x + 3})^2 – 2^2}{(x – 1)(\sqrt{x + 3} + 2)} \]
Nolofatsa palo:
\[ = \frac{x + 3 – 4}{(x – 1)(\sqrt{x + 3} + 2)} \]
\[ = \frac{x – 1}{(x – 1)(\sqrt{x + 3} + 2)} \]
Hlakola \( x – 1 \) (ho tloha \( x \neq 1 \)):
\[ = \frac{1}{\sqrt{x + 3} + 2} \]
Jwale nka sebaka sa \( x = 1 \):
\[ \lim_{x \ho 1} \frac{1}{\sqrt{x + 3} + 2} = \frac{1}{\sqrt{1 + 3} + 2} \]
\[ = \frac{1}{\sqrt{4} + 2} \]
\[ = \frac{1}{2 + 2} \]
\[ = \frac{1}{4} \]
Kahoo, \( \lim_{x \to 1} \frac{\sqrt{x + 3} – 2}{x – 1} = \frac{1}{4} \).
Qetello
Ho utloisisa meeli ea mesebetsi ea algebra ho kenyelletsa mekhoa e fapaneng e kang ho nkeloa sebaka ka ho toba, ho etsa hore lintho li be bonolo, le ho rala mabaka. Ka ho tseba mekhoa ena hantle, re ka sebetsana le mefuta e fapaneng ea mathata a moeli ho calculus. Ha u tobane le mosebetsi o sa tsejoeng, kamehla batla litsela tsa ho nolofatsa mosebetsi e le hore moeli o ka baloa ka nepo. Re tšepa hore mehlala ea mathata le puisano e kaholimo li u thusitse ho utloisisa mohopolo ona hamolemo.
