Mehlala ea Lipotso tse Buisanang ka Meeli ea Mesebetsi ea Trigonometric
Pendahuluan
Moeli oa mosebetsi ke mohopolo oa motheo ho calculus, o hlalosang boleng boo mosebetsi o bo atamelang ha phetoho ea oona e atamela boleng bo itseng. Puisanong ena, re tla tsepamisa maikutlo meeling ea mesebetsi ea trigonometric, e atisang ho hlaha lits'ebetsong tse fapaneng tsa lipalo, ho kenyeletsoa fisiks, boenjiniere le saense ea khomphutha.
Mesebetsi ea Trigonometric joalo ka sin(x), cos(x), le tan(x) e na le litšobotsi tse ikhethang tse etsang hore lipalo tsa eona li be monate haholo. Sehlooho sena se tla tšohla mehlala e 'maloa ea mathata a amanang le meeli ea mesebetsi ea trigonometric, hammoho le litlhaloso tse qaqileng.
Mohlala oa Potso ea 1: Moeli oa Sine
Potso:
Bala moedi \(\lim_{{x \to 0}} \frac{{\sin x}}{x}\).
Puisano:
Moeli ona ke o mong oa meeli ea motheo ho trigonometry 'me o sebelisoa khafetsa ho bopaki le litheoremong tse fapaneng tsa lipalo. Re ka sebelisa Molao oa L'Hôpital kapa tlhaloso ea moeli ho rarolla bothata bona.
Ho Sebelisa Tlhaloso ea Moeli:
Hoa tsebahala hore \( \sin x \approx x \approx x \) jwalo ka \( x \) e atamela 0 (ho sebediswa kgakanyo ya Taylor). Ka hona,
\[
\lim_{{x \to 0}} \frac{{\sin x}}{x} = \lim_{{x \to 0}} \frac{x}{x} = 1.
\]
Ho sebelisa Molao oa L'Hopital:
Kaha sebopeho sa moedi ona ke \(\frac{0}{0}\), re ka sebedisa Molao wa L'Hopital ka ho kgetholla nomoro le denominator.
\[
\lim_{{x \to 0}} \frac{{\sin x}}{x} = \lim_{{x \to 0}} \frac{{\frac{d}{dx} (\sin x)}}{{\frac{d}{dx} (x)}} = \lim_{{x \to 0}} \frac{{\cos x}}{1} = \cos(0) = 1.
\]
Kahoo, sephetho ke 1.
Mohlala Potso ea 2: Moeli oa Cosine
Potso:
Bala moedi \(\lim_{{x \to 0}} \frac{1 – \cos x}{x^2}\).
Puisano:
Ho rarolla moedi ona, re ka sebedisa boitsebiso ba trigonometric kapa mokgwa o tobileng wa ho sebedisa Molao wa L'Hopital.
Ho Sebelisa Boitsebiso ba Trigonometric:
Re hopola boitsebiso boo:
\[ 1 – \cos x = 2 \sin^2 \left( \frac{x}{2} \right). \]
Kahoo moedi o fetoha:
\[
\lim_{{x \to 0}} \frac{1 – \cos x}{x^2} = \lim_{{x \to 0}} \frac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2}.
\]
Ka ho nkela sebaka \( u = \frac{x}{2} \), ebe \( x = 2u \) mme moedi o fetoha ho:
\[
\lim_{{u \to 0}} \frac{2 \sin^2(u)}{(2u)^2} = \lim_{{u \to 0}} \frac{2 \sin^2(u)}{4u^2} = \frac{1}{2} \lim_{{u \to 0}} \left( \frac{\sin u}{u} \right)^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2}.
\]
Ho sebelisa Molao oa L'Hopital:
Sebopeho ke \(\frac{0}{0}\), kahoo re ka sebelisa Molao oa L'Hopital:
\[
\lim_{{x \to 0}} \frac{1 – \cos x}{x^2} = \lim_{{x \to 0}} \frac{\sin x}{2x} = \lim_{{x \to 0}} \frac{\cos x}{2} = \frac{\cos 0}{2} = \frac{1}{2}.
\]
Kahoo, sephetho ke \( \frac{1}{2} \).
Mohlala oa Potso ea 3: Moeli oa Tangent
Potso:
Bala moedi \(\lim_{{x \to 0}} \frac{\tan x}{x}\).
Puisano:
Foromo ena e na le ts'ebetso \(\frac{\sin x}{\cos x}\), 'me e hloka tšebeliso ea meeli ea motheo eo re buileng ka eona pejana.
\[
\lim_{{x \to 0}} \frac{\tan x}{x} = \lim_{{x \to 0}} \frac{\sin x / \cos x}{x} = \lim_{{x \to 0}} \frac{\sin x}{x} \cdot \frac{1}{\cos x}
\]
Re tseba ho tsoa moeling oa motheo hore:
\[
\lim_{{x \to 0}} \frac{\sin x}{x} = 1 \quad \text{and} \quad \lim_{{x \to 0}} \frac{1}{\cos x} = \frac{1}{\cos 0} = 1.
\]
Kahoo, sephetho ke:
\[
1 \ cdot 1 = 1.
\]
Sephetho ke 1.
Mohlala oa 4: Meeli e Rarahaneng ka Sine le Cosine
Potso:
Bala moedi \(\lim_{{x \to 0}} \frac{\sin(2x)}{\cos(3x) – 1}\).
Puisano:
Sebopeho ke \(\frac{0}{0}\), kahoo re ka sebelisa Molao oa L'Hopital:
\[
\lim_{{x \to 0}} \frac{\sin(2x)}{\cos(3x) – 1} = \lim_{{x \to 0}} \frac{2 \cos(2x)}{-3 \sin(3x)}.
\]
Hape foromo ena ke \(\frac{0}{0}\), kahoo re ka sebedisa Molao wa L'Hopital hape:
\[
= \lim_{{x \to 0}} \frac{-4 \sin(2x)}{-9 \cos(3x)} = \lim_{{x \to 0}} \frac{4 \sin(2x)}{9 \cos(3x)}.
\]
Kaha \(\sin(2x) \approx 2x\) le \(\cos(3x) \approx 1\) ha e ntse e atamela 0:
\[
\frac{4 \cdot 0}{9 \cdot 1} = 0.
\]
Sephetho sa ho qetela ke 0.
Qetello
Ka mehlala e fapaneng e kaholimo, re ka bona kamoo mekhoa e fapaneng e sebelisoang ho bala meeli ea mesebetsi ea trigonometric. Tšebeliso ea boitsebiso ba trigonometric, phetolo, le molao oa L'Hôpital e ka ba thuso haholo ho rarolleng mathata a amanang le moeli.
Kutloisiso e felletseng ea meeli ea motheo e kang \(\lim_{{x \to 0}} \frac{{\sin x}}{x} = 1\) le mokhoa oa ho khetholla khafetsa li bohlokoa haholo ho calculus. Ka ho ikoetlisa ho eketsehileng, baithuti ba tla ba le boiphihlelo bo eketsehileng ba ho sebetsana le mefuta e fapaneng ea mathata a moeli oa ts'ebetso ea trigonometric.
