Ngā Tauira Pātai e Matapaki ana i ngā Herenga o ngā Mahi Ārai
Ko te rohenga o tētahi mahi taurangi he ariā taketake i roto i te tātaitai, e tirotiro ana i te whanonga o tētahi mahi ina tata ana ōna uara taurangi ki tētahi pūwāhi. He mea nui te mārama ki ngā rohenga i roto i ngā tono pāngarau maha, tae atu ki te tātari pāngarau me te whakatauira. Ka whakamāramahia e tēnei tuhinga te ariā o te rohenga o tētahi mahi taurangi mā te whakarato i ētahi tauira rapanga me ō rātou otinga.
Te Ariā Taketake o ngā Herenga o ngā Mahi Ārai
I mua i te urunga atu ki ngā tauira raruraru, me arotake tātou i te ariā taketake o ngā rohe. Ko te rohe o tētahi mahi \( f(x) \) i te whakatata atu o \( x \) ki te uara \( a \) ka tohua e:
\[ \lim_{x \to a} f(x) = L \]
ko te tikanga ka whakatata te uara o \( f(x) \) ki \( L \) i te whakatatanga o \( x \) ki \( a \).
Ngā Pātai Tauira me te Kōrero
Tauira Pātai 1: Te Herenga o ngā Mahi Ārai Māmā
Whakatauhia ngā uara rohe e whai ake nei:
\[ \lim_{x \to 2} (3x + 4) \]
Kōrero:
Mō tētahi mahi rārangi pēnei, ka taea e tātou te whakakapi tika i te uara o \( x \) ki te 2:
\[ \lim_{x \to 2} (3x + 4) = 3(2) + 4 = 6 + 4 = 10 \]
Nō reira, \( \lim_{x \to 2} (3x + 4) = 10 \).
Tauira Pātai 2: Te Rohe o te Mahi Poronomial
Whakatauhia ngā uara rohe e whai ake nei:
\[ \lim_{x \to -1} (x^2 + 2x + 1) \]
Kōrero:
Pērā i te pātai tuatahi, ka taea e tātou te whakakapi tika i te uara o \( x \) ki te -1 i roto i te mahi pūrau:
\[ \lim_{x \to -1} (x^2 + 2x + 1) = (-1)^2 + 2(-1) + 1 \]
\[ = 1 – 2 + 1 \]
\[ = 0 \]
Nō reira, \( \lim_{x \to -1} (x^2 + 2x + 1) = 0 \).
Tauira Pātai 3: Te Herenga o ngā Mahi Ārai me ngā Hautau
Whakatauhia ngā uara rohe e whai ake nei:
\[ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} \]
Kōrero:
Ki te whakakapia tika te \( x = 3 \) ki roto i te mahi, ka whiwhi tātou i te āhua kore-taurite \( \frac{0}{0} \). Hei whakaoti i tēnei, me whakarea e tātou:
\[ \frac{x^2 – 9}{x – 3} = \frac{(x – 3)(x + 3)}{x – 3} \]
I mua i te whakakorenga \( x – 3 \), kia mōhio koe ko \( x \neq 3 \), kia taea ai e tātou te whakakore \( x – 3 \):
\[ = x + 3 \]
Whakakapia inaianei \( x = 3 \):
\[ \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 3 + 3 = 6 \]
Nō reira, \( \lim_{x \to 3} \frac{x^2 – 9}{x – 3} = 6 \).
Tauira Raru 4: Ngā Herenga o ngā Mahi me ngā Pūtake
Whakatauhia ngā uara rohe e whai ake nei:
\[ \lim_{x \to 4} \sqrt{2x + 1} \]
Kōrero:
Nā te mea he mahi tonu te mahi i roto i ngā pūtake, ka taea e tātou te whakakapi tika i te uara o \( x = 4 \):
\[ \lim_{x \to 4} \sqrt{2x + 1} = \sqrt{2(4) + 1} \]
\[ = \sqrt{8 + 1} \]
\[ = \sqrt{9} \]
\[ = 3 \]
Nō reira, \( \lim_{x \to 4} \sqrt{2x + 1} = 3 \).
Tauira Pātai 5: Te Herenga o ngā Mahi Ārai me te Whakamārama
Whakatauhia ngā uara rohe e whai ake nei:
\[ \lim_{x \to 1} \frac{\sqrt{x + 3} – 2}{x – 1} \]
Kōrero:
Mā te whakakapinga tika \( x = 1 \) ka puta te āhua kore-tau \( \frac{0}{0} \). Nō reira, me whakamārama tātou. Whakareatia te taupū me te tauwehe ki ō rāua takirua e rite ana:
\[ \frac{\sqrt{x + 3} – 2}{x – 1} \times \frac{\sqrt{x + 3} + 2}{\sqrt{x + 3} + 2} = \frac{(\sqrt{x + 3})^2 – 2^2}{(x – 1)(\sqrt{x + 3} + 2)} \]
Whakangāwaritia te taunga:
\[ = \frac{x + 3 – 4}{(x – 1)(\sqrt{x + 3} + 2)} \]
\[ = \frac{x – 1}{(x – 1)(\sqrt{x + 3} + 2)} \]
Whakakore \( x – 1 \) (mai i \( x \neq 1 \)):
\[ = \frac{1}{\sqrt{x + 3} + 2} \]
Whakakapia inaianei \( x = 1 \):
\[ \lim_{x \to 1} \frac{1}{\sqrt{x + 3} + 2} = \frac{1}{\sqrt{1 + 3} + 2} \]
\[ = \frac{1}{\sqrt{4} + 2} \]
\[ = \frac{1}{2 + 2} \]
\[ = \frac{1}{4} \]
Nō reira, \( \lim_{x \to 1} \frac{\sqrt{x + 3} – 2}{x – 1} = \frac{1}{4} \).
Whakamutunga
Ko te mārama ki ngā rohe o ngā mahi taurangi he maha ngā tikanga pēnei i te whakakapinga tika, te whakawehewehe, me te whakamārama. Mā te mōhio ki ēnei tikanga, ka taea e tātou te whakatau i ngā momo raruraru rohe i roto i te tātaitai. Ina tūtaki ki tētahi mahi kore-taurite, rapua tonu ngā huarahi hei whakahaere i te mahi kia taea ai te tatau tika i te rohe. Ko te tumanako, kua āwhina ngā tauira raruraru me te matapakinga i runga ake nei i a koe ki te mārama ake ki tēnei ariā.
