Ngā Rohe o ngā Mahi Pāngatoru
He ariā taketake ngā rohe i roto i te tātaitai e puta ana i roto i ngā manga maha o te pāngarau me te pūtaiao. He taputapu tino whai hua ngā rohe i roto i te tātari i ngā mahi me ngā huringa, tae atu ki te mārama ki te whanonga o ngā mahi pākoki i a rātou e whakatata atu ana ki tētahi wāhi. I roto i tēnei tuhinga, ka tūhuratia e mātou te ariā o ngā rohe i roto i te horopaki o ngā mahi pākoki, tae atu ki ngā tikanga mō te tatau i ngā rohe me ngā tauira.
Te Whakamāramatanga o te Here
I ngā kupu māmā noa iho, ko te rohe he uara e whakatata atu ana tētahi mahi ina whakatata atu tōna taurangi motuhake ki tētahi uara. Hei tauira, mēnā he mahi tā tātou \( f(x) \), ko te rohe o \( f(x) \) ina whakatata atu a \( x \) ki \( a \) ka whakaaturia penei:
\[ \lim_{x \to a} f(x) = L \]
Ko te tikanga tēnei, ka tata haere a \( x \) ki \( a \), ka tata haere a \( f(x) \) ki \( L \).
Ngā Mahi me ngā Herenga o te Pāngatoru
He whānuitia te whakamahinga o ngā mahi taurite pēnei i te sine (sin), te cosine (cos), te tangent (tan), me te secant (sec) i roto i ngā momo mahi. He mea nui te mārama ki ngā rohe o ēnei mahi i roto i te tātaritanga pāngarau me te whakatauira.
Ngā Herenga Taketake o ngā Mahi Pāngatoru
Me tīmata tātou me ētahi rohe taketake e puta pinepine ana i roto i te tātaitanga pākoki:
1. Te rohenga o te mahi Sine:
\[ \lim_{x \to 0} \sin(x) = 0 \]
2. Te rohenga o te mahi kōsina:
\[ \lim_{x \to 0} \cos(x) = 1 \]
3. Te rohenga o te mahi pātata:
\[ \lim_{x \to 0} \tan(x) = 0 \]
He mea tino nui te whakawhāiti i te kore i roto i te ine pākoki, nā te mea he maha ngā ariā me ngā tuakiri pākoki e hangai ana ki te whanonga o tēnei mahi huri noa i te kore.
Ngā Herenga Taketake o te Ine Taurite
He maha ngā herenga motuhake e pā ana ki ngā mahi pākoki, ā, e whakamahia pinepinetia ana i roto i te tātaitai. Hei tauira:
1. Te rohenga o te Sine mō ia x:
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]
2. Tepe 1 – Kohinu mō ia x^2:
\[ \lim_{x \to 0} \frac{1 – \cos(x)}{x^2} = \frac{1}{2} \]
Ka taea te whakamātau i ēnei rohe mā te whakamahi i tētahi huarahi ā-ira, mā te tikanga rānei a L'Hôpital, e ahu mai ana i ngā tātaitanga.
Te Whakamātautau i ngā Herenga mā te Tikanga a L'Hôpital
He taputapu tino whai hua te tikanga a L'Hôpital mō te tatau i ngā rohe e ahua kore e taea te whakatau mā te whakakapinga tika. Ko te tātai taketake mō te tikanga a L'Hôpital ko:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
me te tikanga \( \lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0 \) me \( \infty / \infty \).
Me whakamahi tātou i tēnei tikanga hei whakamatau i tētahi o ngā herenga matua i runga ake nei:
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]
Ki te whakamātau tātou i te whakakapinga tika, ka puta te āhua \( 0/0 \), kāore i te tautuhia. Mā te whakamahi i te tikanga a L'Hôpital:
\[ f(x) = \sin(x) \kuputuhi{ me } g(x) = x \]
Nā reira:
\[ f'(x) = \cos(x) \text{ me } g'(x) = 1 \]
Muri iho, whakamahia te tikanga a L'Hôpital:
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1 \]
Ngā Tauira o ngā Whakamahinga o ngā Herenga Mahi Pāngatoru
Hei tiro i te mahi a ngā rohe o ngā mahi pākoki i roto i tētahi horopaki uaua ake, me titiro tātou ki ētahi tauira:
Tauira 1: Te rohenga o tētahi mahi whakakotahi
Me kī tātou e hiahia ana ki te tatau i te rohenga e whai ake nei:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} \]
Hei whakaoti i tēnei, ka taea e tātou te whakakapi i te \( u = 2x \), kia \( u \to 0 \) hoki ina \( x \to 0 \). Ko tā tātou rohe ka:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{u \to 0} \frac{\sin(u)}{\frac{u}{2}} = 2 \lim_{u \to 0} \frac{\sin(u)}{u} = 2 \cdot 1 = 2 \]
Tauira 2: Te Whakawhāiti me te Mahi Wehewehe Aho
Whakaarohia ngā rohe e whai ake nei:
\[ \lim_{x \to 0} \frac{1 – \cos(x)}{x^2} \]
E mōhio ana tātou:
\[ \lim_{x \to 0} \frac{1 – \cos(x)}{x^2} = \frac{1}{2} \]
Ka taea te whakaatu anō i tēnei rohe mā te whakamahi i te tikanga a L'Hôpital nā te mea ina whakakapia tika, ka puta te āhua \( 0/0 \):
\[ f(x) = 1 – \cos(x) \kuputuhi{ me } g(x) = x^2 \]
Ko ngā pānga tuatahi o ēnei mahi ko:
\[ f'(x) = \sin(x) \text{ me } g'(x) = 2x \]
Nā, mā te tikanga a L'Hôpital:
\[ \lim_{x \to 0} \frac{1 – \cos(x)}{x^2} = \lim_{x \to 0} \frac{\sin(x)}{2x} = \frac{1}{2} \lim_{x \to 0} \frac{\sin(x)}{x} = \frac{1}{2} \cdot 1 = \frac{1}{2} \]
Whakamutunga
Ko te mārama ki ngā rohe o ngā mahi pākoki he tūāpapa pakari mō ngā ariā uaua ake i roto i te tātaitai me te tātari pāngarau. Ko ngā rohe pēnei i te \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) ehara i te mea ko ngā tuakiri pāngarau anake, engari he taputapu nui hoki e taea ai e tātou te mārama hohonu ki te panoni, te whakatata, me te whanonga o ngā mahi. Mā te mōhio ki ēnei ariā, ka taea e tātou te tātari pai ake i ngā āhuatanga taiao me ngā tono hangarau maha e hangai ana ki te pāngarau.
