Soal Limit Trigonometri Sbmptn
soal limit trigonometri
1. soal limit trigonometri
Penjelasan dengan langkah-langkah:
[tex]\lim_{x \to 0}( \frac{ \sin(2x) \cos(3x)}{5x} ) \\ [/tex]
Menggunakan aturan L'Hopital[tex] = \frac{2 \cos(2 \times 0) \cos(3 \times 0) - 3 \sin(2 \times 0) \sin(3 \times 0)}{5} [/tex]
[tex] = \frac{2 \cos(0) \cos(0) - 3 \sin(0) \sin(0)}{5} [/tex]
[tex] = \frac{2 \times 1 \times 1 - 3 \times 0 \times 0}{?} [/tex]
[tex] = \frac{2 - 0}{5} [/tex]
[tex]{ \boxed{ \boxed{ \rm = \frac{2}{5} }}}[/tex]
#BudayakanBerterimaKasih :)
2. soal limit trigonometri
Jawab:
Penjelasan dengan langkah-langkah:
limit trigonometri
ljm x→0 sin ax/bx = lim x→0 sin ax/bx = a/b
•
lim x→0 (1 - cos² x)/(x tan 2x)
= lim x→0 sin² x / x tan 2x
= lim x→0 (sin x/x) . (sin x/tan 2x)
= x/x . x/2x
= 1 . 1/2
= 1/2
3. soal limit trigonometri
semoga bermanfaat ya [tex]\lim_{x\to0}{\frac{x+\sin{2x}}{2x-\tan{6x}}}=\\[/tex]
Bentuk ini bisa diselesaikan dengan manipulasi aljabar, yaitu dengan menambahkan bentuk [tex]\frac{\frac{1}{x}}{\frac{1}{x}}[/tex], sehingga :
[tex]\lim_{x\to0}{\frac{x+\sin{2x}}{2x-\tan{6x}}}=\lim_{x\to0}{\frac{x+\sin{2x}}{2x-\tan{6x}}.\frac{\frac{1}{x}}{\frac{1}{x}}}\\\lim_{x\to0}{\frac{\frac{x+\sin{2x}}{x}}{\frac{2x-\tan{6x}}{x}}}=\frac{1+\lim_{x\to0}{\frac{\sin{2x}}{x}}}{2-\lim_{x\to0}{\frac{\tan{6x}}{x}}}=\frac{1+\lim_{x\to0}{\frac{\sin{2x}}{2x}.2}}{2-\lim_{x\to0}{\frac{\tan{6x}}{6x}.6}}=\frac{1+1.2}{2-1.6}=\frac{3}{-4}=-\frac{3}{4}\\[/tex]
Semoga membantu.
4. Matematika soal limit trigonometri
langsung aja ya
Lim (cosx / x sinx - cos^2x/xsinx)
Lim (cos x - cos^2x)/xsinx
Lim cosx(1 - cosx)/xsinx
Lim cosx . 2.sin^2(1/2x) / xsinx
2.Lim cosx . Lim sin(1/2x)/x . Lim sin(1/2x)/sinx
2.cos 0° . (1/2)/1 . (1/2)/1
2.1.1/2.1/2
2/4
1/2
Jwb. E
*Lim x-->0
* cos2x = 1 - sin^2x
2sin^2x = 1 - cos2x
2sin^2(1/2x) = 1 - cosx
5. soal limit trigonometri
semogamembantu^_^semangattrusbelajarnya^_^enjoyyy
6. soal limit trigonometri
[tex]\lim_{x \to 0}\: \left( \frac{2x - x\sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \: cos \: 4x}}}}{tan \: x \: - \: sin \: x} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{2x - x\sqrt{2 + \sqrt{2 + \sqrt{2 + 2(2 \: {cos}^{2} \:2x - 1)}}}}{ \frac{sin \: x}{cos \: x} \: - \: sin \: x} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{2x - x\sqrt{2 + \sqrt{2 + \sqrt{2 + 4 \: {cos}^{2} \:2x - 2}}}}{ \frac{sin \: x - sin \: x \: cosx}{cos \: x} } \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x \left(2x - x\sqrt{2 + \sqrt{2 + 2 \: cos \: 2x}} \right)}{ sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{ cos \: x\left(2x - x\sqrt{2 + \sqrt{2 + 2(2\: {cos}^{2} \: x - 1)}} \right)}{sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x \left(2x - x\sqrt{2 + \sqrt{2 + 4\: {cos}^{2} \: x - 2}} \right)}{sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x \left(2x - x\sqrt{2 + 2 \: cos \: x} \right)}{sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x \left(2x - x\sqrt{2 + 2(2\: {cos}^{2} \: \frac{1}{2} x - 1)} \right)}{sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x \left(2x - x\sqrt{2 + 4\: {cos}^{2} \: \frac{1}{2} x - 2} \right)}{sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x(2x - x(2 \: cos \: \frac{1}{2} x))}{sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x(2x(1 - cos \: \frac{1}{2} x))}{sin \: x(1 - cos \: x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{cos \: x(2x(2 \: {sin}^{2} \: x))}{sin \: x(2 \: {sin}^{2} \: \frac{1}{2}x)} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{x(2 \: cos \: x\: sin \: x)}{{sin}^{2} \: \frac{1}{2}x} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{(2)( \frac{1}{2}x)(sin \: 2x)}{{sin}^{2} \: \frac{1}{2}x} \right)[/tex]
[tex]= \lim_{x \to 0}\: \left( \frac{ \frac{1}{2}x}{sin \: \frac{1}{2}x} \right).\left( \frac{(2)(sin \: 2x)}{sin\: \frac{1}{2}x} \right)[/tex]
[tex]= \lim_{x \to 0}\: (2).\left( \frac{ \frac{1}{2}x}{sin \: \frac{1}{2}x} \right).\left( \frac{sin \: 2x}{sin\: \frac{1}{2}x} \right)[/tex]
[tex]= (2).(1).\left( \frac{2}{\frac{1}{2}} \right)[/tex]
[tex] \boxed{ \boxed{ = 8}}[/tex]
7. Soal limit trigonometri
Jawab:
Penjelasan dengan langkah-langkah:
8. hasil Limit Trigonometri dari soal ini adalah
Limit
lim (x→0) (1 - cos³ x) / sin² x
L'Hos
= lim (x→0) (3 cos² x sin x) / (2 sin x cos x)
= lim (x→0) (3 cos x)/2
= 3/2 cos 0
= 3/2 ✔
9. Soal Limit trigonometri....
Jawab
34.
soal
= lim x→π/4 (cos² x - sin² x) . cos x / (cos x - sin x)
= lim x→π/4 cos x (cos x + sin x)(cos x - sin x) / (cos x - sin x)
= lim x→π/4 cos² x + cos x sin x
= (1/2 √2)² + 1/2 √2 . 1/2 √2
= 2/4 + 2/4
= 1
25.
soal
= lim x→π/2 sin² (π/2 - x) / 2(x - π/2) sin (x - π/2)
= lim x→π/2 (- sin (x - π/2))² / 2(x - π/2) sin (x - π/2)
= lim x→π/2 sin (x - π/2) / 2(x - π/2)
= 1/2
10. soal limit trigonometri
limit trigonometri
lim x→ (cos x - cos 5x) / (cos x tan² 2x)
= lim x→0 (-2 sin (x + 5x)/2 sin (x - 5x)/2) /(cos x tan² 2x)
= lim x→0 (2 sin 3x sin 2x) / cos x tan² 2x
= lim x→0 2(sin 3x/tan 2x) . (sin 2x/tan 2x) . (1/cos x)
= 2 . 3x/2x . 2x/2x . 1/cos 0
= 2 . 3/2 . 1 . 1
= 3
11. soal tentang limit trigonometri..
Nilai dari [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{1 - cos \: x}{sin \: 3x \: tan \: 2x}}}[/tex] adalah [tex]\boxed{\sf{\dfrac{1}{12}}}.[/tex]
ㅤPEMBAHASANLimit fungsi merupakan keadaan dari suatu fungsi saat mendekati suatu titik. Misalnya fungsi f(x) tidak terdefinisi saat x = a namun bernilai L saat mendekati a. Secara matematis dapat dituliskan menjadi:
[tex]\boxed{\boxed{\sf{\lim_{x \to a}f(x) = L}}}[/tex]
ㅤTeorema LimitBerikut beberapa teorema limit utama.
→ [tex]\displaystyle{\sf{\lim_{x \to a}k = k}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a} {k. \: x}^{n} = k. \: {a}^{n}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}k. \: f(x) = k. \: \lim_{x \to a} \: f(x)}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}f(x) \pm g(x) = \lim_{x \to a}f(x) \pm\lim_{x \to a}g(x)}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}f(x) \times g(x) = \lim_{x \to a}f(x) \times \lim_{x \to a}g(x)}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\displaystyle{\sf{\lim_{x \to a}f(x)}}}{\displaystyle{\sf{\lim_{x \to a}g(x)}}}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}{\left[f(x)\right]}^{n} = {\left[\lim_{x \to a}f(x)\right]}^{n}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}\sf{\sqrt[\sf{n}]{\sf{f(x)}}} = \sqrt[\sf{n}]{\displaystyle{\sf{\lim_{x \to a}f(x)}}}}}[/tex]
ㅤBerikut beberapa teorema limit trigonometri.
→ [tex]\displaystyle{\sf{\lim_{x \to a}sin \: x = sin \: a}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}cos \: x = cos \: a}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}tan \: x = tan \: a}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}csc \: x = csc \: a}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}sec\: x = sec \: a}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to a}cot \: x = cot \: a}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{sin \: x}{x} = 1}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{x}{sin \: x} = 1}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{tan \: x}{x} = 1}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{x}{tan \: x} = 1}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{sin \: ax}{bx} = \dfrac{a}{b}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{ax}{sin \: bx} = \dfrac{a}{b}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{tan \: ax}{bx} = \dfrac{a}{b}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{ax}{tan \: bx} = \dfrac{a}{b}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{sin \: ax}{sin \: bx} = \dfrac{a}{b}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{tan \: ax}{tan \: bx} = \dfrac{a}{b}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{sin \: ax}{tan \: bx} = \dfrac{a}{b}}}[/tex]
→ [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{tan \: ax}{sin \: bx} = \dfrac{a}{b}}}[/tex]
ㅤIngat:
[tex]\boxed{\boxed{\sf{cos \: ax} = \left\{\begin{array}{c} \sf{{cos}^{2}\dfrac{a}{2}x - {sin}^{2} \dfrac{a}{2}x}\\ \\ \sf{2 \: {cos}^{2}\dfrac{a}{2}x - 1} \\ \\\sf{1 - 2 \: {sin}^{2}\dfrac{a}{2}x}\end{array}\right.}}[/tex]
ㅤDiketahui:
[tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{1 - cos \: x}{sin \: 3x \: tan \: 2x}}}[/tex]
ㅤDitanyakan:
Nilai dari [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{1 - cos \: x}{sin \: 3x \: tan \: 2x}}}[/tex]
ㅤJawab:
[tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{1 - cos \: x}{sin \: 3x \: tan \: 2x} = \lim_{x \to 0}\dfrac{1 -(1 - 2 \: {sin}^{2}\tfrac{1}{2}x)}{sin \: 3x \: tan \: 2x}}} \\ \\ \displaystyle{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\sf{ = \lim_{x \to 0} \dfrac{\cancel{1} - \cancel{1} + {2 \: sin}^{2}\tfrac{1}{2}x}{sin \: 3x \: tan \: 2x}}}\\ \\\displaystyle{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\sf{ = \lim_{x \to 0}\dfrac{{2 \: sin}^{2}\tfrac{1}{2}x}{sin \: 3x \: tan \: 2x}}}\\ \\\displaystyle{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\sf{ = \lim_{x \to 0}2. \: \lim_{x \to 0}\dfrac{\: \: sin \: \tfrac{1}{2}x \: \: }{sin \: 3x}. \: \lim_{x \to 0}\dfrac{\: \: sin \: \tfrac{1}{2}x \: \: }{tan \: 2x}}} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:{ = 2. \: \dfrac{ \: \: \tfrac{1}{2} \: \: }{3}. \: \dfrac{ \: \: \tfrac{1}{2} \: \: }{2}}} \\ \\ \sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 2.\:\dfrac{1}{6}.\:\dfrac{1}{4}} \\ \\ \sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: = \dfrac{1}{12}}[/tex]
ㅤJadi nilai dari [tex]\displaystyle{\sf{\lim_{x \to 0} \dfrac{1 - cos \: x}{sin \: 3x \: tan \: 2x}}}[/tex] adalah [tex]\boxed{\sf{\dfrac{1}{12}}}.[/tex]
ㅤPELAJARILEBIHLANJUTKasus limit trigonometri lainnya dapat disimak juga di:
brainly.co.id/tugas/24724411brainly.co.id/tugas/23465822brainly.co.id/tugas/30234598ㅤDETAILJAWABANKelas:11
Mapel:Matematika
Materi:Limit Fungsi
KodeKategorisasi:11.2.8
KataKunci:Limit Fungsi, Limit Trigonometri, Teorema Limit Utama, Teorema Limit Trigonometri, Limit Fungsi Trigonometri Menuju Nol
12. Soal Matematika Limit Trigonometri.
lim x-->0 {√(x+x²) - √x}/x√x
= lim x-->0 √x(√(1+x) - 1) / x√x
= lim x -->0 (√(1+x) - 1)/x
= lim x -->0 (√(1+x) - 1)/x * (√(1+x) +1)/(√(1+x) +1)
= lim x--> 0 1+x-1 / (x * (√(1+x) +1))
= lim x-->0 x /( x * (√(1+x) +1))
= lim x--> 0 1/(√(1+x) +1)
= 1/(√1 +1) = 1/2 (D)Lim (√(x + x^2) - √x) / x√x . (√(x + x^2) + √x)/(√(x + x^2) + √x)
= Lim (x + x^2 - x) / [x√x . (√(x(1 + x)) + √x)]
= Lim x^2 / [x√x . √x (√(1 + x) + 1)]
= Lim x^2 / [x^2 (√(1 + x) + 1)]
= Lim 1/(√(1 + x) + 1)
= 1/(√(1 + 0) + 1)
= 1/2
13. soal limit fungsi trigonometri
semoga membantu......
14. soal limit fungsi trigonometri
Jawab:
Penjelasan dengan langkah-langkah:
15. Latihan soal limit trigonometri
mksd nya ap rusuh buat soal y
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