Ngā Tauira Pātai e Matapaki ana i ngā Herenga o ngā Mahi Pāngatoru
Pendahuluan
Ko te rohe o tētahi mahi he ariā taketake i roto i te tātaitai, e whakaahua ana i te uara e whakatata atu ana tētahi mahi ina whakatata atu tōna taurangi ki tētahi uara. I roto i tēnei kōrero, ka arotahi tātou ki ngā rohe o ngā mahi pākoki, e puta pinepine ana i roto i ngā tono maha o te pāngarau, tae atu ki te ahupūngao, te hangarau, me te pūtaiao rorohiko.
He āhuatanga ahurei ō ngā mahi whārite pēnei i te sin(x), te cos(x), me te tan(x) e whakamere ai ā rātou tataunga. Ka matapakihia e tēnei tuhinga ētahi tauira raruraru e pā ana ki ngā rohe o ngā mahi whārite, me ngā whakamārama taipitopito.
Tauira Pātai 1: Te Rohe o te Sine
Pātai:
Tātaihia te rohe \(\lim_{{x \to 0}} \frac{{\sin x}}{x}\).
Kōrero:
Ko tēnei rohe tētahi o ngā rohe taketake o te ine whārite, ā, e whakamahia pinepinetia ana i roto i ngā momo taunakitanga me ngā ariā i roto i te tātaitai. Ka taea e tātou te whakamahi i te Ture a L'Hôpital, i te whakamāramatanga rānei o te rohe hei whakaoti i tēnei raruraru.
Te Whakamahi i te Whakamāramatanga Rohe:
E mōhiotia ana ko \( \sin x \approx x \) i te whakatata atu o \( x \) ki te 0 (mā te whakamahi i te whakatau tata a Taylor). Nō reira,
\[
\lim_{{x \to 0}} \frac{{\sin x}}{x} = \lim_{{x \to 0}} \frac{x}{x} = 1.
\]
Mā te whakamahi i te Ture a L'Hopital:
Nā te mea ko te āhua o tēnei rohe ko \(\frac{0}{0}\), ka taea e tātou te whakamahi i te Ture a L'Hopital mā te wehewehe i te taupū me te tauwehe.
\[
\lim_{{x \to 0}} \frac{{\sin x}}{x} = \lim_{{x \to 0}} \frac{{\frac{d}{dx} (\sin x)}}{{\frac{d}{dx} (x)}} = \lim_{{x \to 0}} \frac{{\cos x}}{1} = \cos(0) = 1.
\]
Nō reira, ko te hua ko te 1.
Tauira Pātai 2: Te rohenga o te Cosine
Pātai:
Tātaihia te rohenga \(\lim_{{x \to 0}} \frac{1 – \cos x}{x^2}\).
Kōrero:
Hei whakaoti i tēnei rohe, ka taea e tātou te whakamahi i ngā tuakiri pākoki, i tētahi huarahi tika rānei mā te Ture a L'Hopital.
Te Whakamahi i ngā Tuakiri Pātoru:
Ka maumahara tātou ki te tuakiri e:
\[ 1 – \cos x = 2 \sin^2 \left( \frac{x}{2} \right). \]
Nō reira ka noho te rohe:
\[
\lim_{{x \to 0}} \frac{1 – \cos x}{x^2} = \lim_{{x \to 0}} \frac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2}.
\]
Mā te whakakapinga \( u = \frac{x}{2} \), kātahi ka \( x = 2u \) ka huri te rohe ki:
\[
\lim_{{u \to 0}} \frac{2 \sin^2(u)}{(2u)^2} = \lim_{{u \to 0}} \frac{2 \sin^2(u)}{4u^2} = \frac{1}{2} \lim_{{u \to 0}} \left( \frac{\sin u}{u} \right)^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2}.
\]
Mā te whakamahi i te Ture a L'Hopital:
Ko te āhua ko \(\frac{0}{0}\), nō reira ka taea e tātou te whakamahi i te Ture a L'Hopital:
\[
\lim_{{x \to 0}} \frac{1 – \cos x}{x^2} = \lim_{{x \to 0}} \frac{\sin x}{2x} = \lim_{{x \to 0}} \frac{\cos x}{2} = \frac{\cos 0}{2} = \frac{1}{2}.
\]
Nō reira, ko te hua ko \( \frac{1}{2} \).
Tauira Pātai 3: Tepe Pānga
Pātai:
Tātaihia te rohenga \(\lim_{{x \to 0}} \frac{\tan x}{x}\).
Kōrero:
Kei roto i tēnei puka te mahi \(\frac{\sin x}{\cos x}\), ā, me whakamahi i ngā rohenga taketake i kōrerotia i mua ake nei.
\[
\lim_{{x \to 0}} \frac{\tan x}{x} = \lim_{{x \to 0}} \frac{\sin x / \cos x}{x} = \lim_{{x \to 0}} \frac{\sin x}{x} \cdot \frac{1}{\cos x}
\]
E mōhio ana tātou mai i te rohe taketake:
\[
\lim_{{x \to 0}} \frac{\sin x}{x} = 1 \quad \text{and} \quad \lim_{{x \to 0}} \frac{1}{\cos x} = \frac{1}{\cos 0} = 1.
\]
Nā, ko te hua:
\[
1 \cdot 1 = 1.
\]
Ko te hua ko te 1.
Tauira 4: Ngā Herenga Uaua me te Sine me te Cosine
Pātai:
Tātaihia te rohe \(\lim_{{x \to 0}} \frac{\sin(2x)}{\cos(3x) – 1}\).
Kōrero:
Ko te āhua ko \(\frac{0}{0}\), nō reira ka taea e tātou te whakamahi i te Ture a L'Hopital:
\[
\lim_{{x \to 0}} \frac{\sin(2x)}{\cos(3x) – 1} = \lim_{{x \to 0}} \frac{2 \cos(2x)}{-3 \sin(3x)}.
\]
Ko tēnei puka anō ko \(\frac{0}{0}\), nō reira ka taea e tātou te whakamahi anō i te Ture a L'Hopital:
\[
= \lim_{{x \to 0}} \frac{-4 \sin(2x)}{-9 \cos(3x)} = \lim_{{x \to 0}} \frac{4 \sin(2x)}{9 \cos(3x)}.
\]
Mai i te mea ko \(\sin(2x) \approx 2x\) me \(\cos(3x) \approx 1\) e whakatata atu ana ki te 0:
\[
\frac{4 \cdot 0}{9 \cdot 1} = 0.
\]
Ko te hua whakamutunga ko te 0.
Whakamutunga
Mā roto i ngā tauira rerekē i runga ake nei, ka kitea te whakamahinga o ngā huarahi rerekē hei tatau i ngā rohe o ngā mahi pākoki. Ka tino āwhina te whakamahinga o ngā tuakiri pākoki, te whakakapinga, me te ture a L'Hôpital ki te whakaoti rapanga e pā ana ki te rohe.
He mea nui te māramatanga hōhonu ki ngā rohenga taketake pēnei i te \(\lim_{{x \to 0}} \frac{{\sin x}}{x} = 1\) me te tikanga o te rerekētanga auau i roto i te tātaitai. Mā te whakaharatau tonu, ka matatau ake ngā ākonga ki te whakaoti i ngā momo raruraru rohenga mahi pākoki.
