Imagine a particle of mass m, constrained to move along the x-axis, subject to some specified force F(x, t). The program of classical mechanics is to deter- mine the position of the particle at any given time: x(t). Once we know that, we can figure out the velocity (\( v=\frac{dx}{dt}\) ), the momentum (p = mv), the kinetic energy ( \( T=\frac{1}{2}mv^2 \) ), or any other dynamical variable of interest. And how do we go about determining x(t)? We apply Newton's second law: F = ma. (For conservative systems the only kind we shall consider, and, fortunately, the only kind that occur at the microscopic level---the force can be expressed as the derivative of a potential energy function, \( F=-\frac{\partial V}{\partial x} \) , and Newton's law reads \( m\frac{d^2x}{dt^2}=-\frac{\partial V}{\partial x} \) .) This, together with appropriate initial conditions (typically the position and velocity at t 0), determines x(t). Quantum mechanics approaches this same problem quite differentl...
Soal dan Pembahasan Limit Fungsi Trigonometri - Limit fungsi trigonometri adalah salah satu materi yang dipelajari dalam mata pelajaran Matematika, terutama pada jenjang pendidikan menengah atas. Limit fungsi trigonometri sendiri merujuk pada batasan nilai dari suatu fungsi trigonometri ketika variabel x mendekati suatu nilai tertentu. Untuk memahami limit fungsi trigonometri, kita perlu menguasai terlebih dahulu konsep limit pada umumnya.
Beberapa contoh soal dan pembahasan limit fungsi trigonometri meliputi kelas 12 dan UTBK. Dalam pembahasan tersebut, seringkali diberikan contoh soal tentang limit fungsi trigonometri yang harus dipecahkan dengan menggunakan metode-metode tertentu seperti metode substitusi langsung atau teorema limit. Dalam pembahasan limit fungsi trigonometri, biasanya diberikan pula contoh soal dan pembahasan yang berupa file pdf sehingga memudahkan dalam proses belajar.
Salah satu topik yang sering ditekankan dalam pembahasan limit fungsi trigonometri adalah limit tak hingga. Limit tak hingga terjadi ketika variabel x mendekati nilai tak hingga atau minus tak hingga. Selain itu, limit fungsi trigonometri juga sering diasosiasikan dengan fungsi-fungsi trigonometri seperti sinus, kosinus, dan tangen.
Soal Nomor 1. \( \lim_{x \rightarrow 0}\frac{9x - \tan x}{x + \sin 3x} = \) ....
A. -1
B. 0
C. 1
D. 2
E. 3
Pembahasan :
Cara 1 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{9x - \tan x}{x + \sin 3x} &= \lim_{x \rightarrow 0}\frac{9 -\frac{1}{x}\tan x}{1 + \frac{1}{x} \sin 3x} \\
&= \frac{9-1}{1+3} \\
&= 2
\end{align*}
Cara 2 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{9x - \tan x}{x + \sin 3x} &= \lim_{x \rightarrow 0}\frac{9x -x}{x + 3x} \\
&= \frac{8x}{4x} \\
&= 2
\end{align*}
Jawaban : D
Soal Nomor 2. \( \lim_{x \rightarrow 0}\frac{3x - \sin 4x}{5x + \tan 2x} = \) ....
A. \(\frac{3}{5} \)
B. -1
C. 2
D. \(\frac{7}{3} \)
E. ∞
Pembahasan :
Cara 1 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{3x - \sin 4x}{5x + \tan 2x} &= \lim_{x \rightarrow 0}\frac{3 +\frac{1}{x}\sin 4x}{5x - \frac{1}{x}\tan 2x} \\
&= \frac{3+4}{5-2} \\
&= \frac{7}{3}
\end{align*}
Cara 2 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{3x - \sin 4x}{5x + \tan 2x} &= \lim_{x \rightarrow 0}\frac{3x +4x}{5x - 2x} \\
&= \lim_{x \rightarrow 0} \frac{7x}{3x} \\
&= \frac{7}{3}
\end{align*}
Jawaban : D
Soal Nomor 3. \( \lim_{x \rightarrow k}\frac{2k - 2x}{\sin (x-k)+ \tan (3x-3k)} = \) ....
A. -1
B. \(-\frac{1}{2} \)
C. 0
D. \(\frac{1}{2} \)
E. 1
Pembahasan :
\begin{align*}
\lim_{x \rightarrow k}\frac{2k - 2x}{\sin (x-k)+ \tan (3x-3k)} &= \lim_{x \rightarrow k}\frac{2k - 2x}{(x-k)+(3x - 3k)} \\
&= \lim_{x \rightarrow k} \frac{-2(x-k)}{4(x-k)} \\
&= \frac{-2}{4} \\
&= -\frac{1}{2}
\end{align*}
Jawaban : B
Soal Nomor 4. \( \lim_{x \rightarrow a}\frac{7\sin (x-a) +5\tan (x-a)}{x-a + \sin (x-a)} = \) ....
A. 0
B. 2
C. 4
D. 6
E. 8
Pembahasan :
\begin{align*}
\lim_{x \rightarrow a}\frac{7\sin (x-a) +5\tan (x-a)}{x-a + \sin (x-a)} &= \lim_{x \rightarrow a}\frac{7(x-a) +5(x-a)}{x-a + (x-a)} \\
&= \lim_{x \rightarrow a} \frac{12(x-a)}{2(x-a)} \\
&= \frac{12}{2} \\
&= 6
\end{align*}
Jawaban : D
Soal Nomor 5. \( \lim_{x \rightarrow 0}\frac{\sin 4x \tan ^2 3x +6x^3}{2x^2 \sin 3x \cos 2x} = \) ....
A. 0
B. 3
C. 4
D. 5
E. 7
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 0}\frac{\sin 4x \tan ^2 3x +6x^3}{2x^2 \sin 3x \cos 2x} &= \lim_{x \rightarrow 0}\frac{4x (3x)^2 +6x^3}{2x^2 (3x) 1} \\
&= \lim_{x \rightarrow 0} \frac{42x^3}{6x^3} \\
&= \frac{42}{6} \\
&= 7
\end{align*}
Jawaban : E
Soal Nomor 6. \( \lim_{x \rightarrow 0}\frac{\sin (2x^2)}{x^2 +(\sin 3x)^2} = \) ....
A. \(\frac{2}{3} \)
B. 5
C. \(\frac{3}{2} \)
D. 0
E. \(\frac{1}{5} \)
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 0}\frac{\sin (2x^2)}{x^2 +(\sin 3x)^2} &= \lim_{x \rightarrow 0}\frac{2x^2}{x^2 + (3x)^2} \\
&= \lim_{x \rightarrow 0} \frac{3x^2}{10x^2} \\
&= \frac{2}{10} \\
&= \frac{1}{5}
\end{align*}
Jawaban : E
Soal Nomor 7. \( \lim_{x \rightarrow 0}\frac{(x^2 -1)\sin 6x}{x^3 +3x^2 +2} = \) ....
A. -3
B. -1
C. 0
D. 1
E. 6
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 0}\frac{(x^2 -1)\sin 6x}{x^3 +3x^2 +2} &= \lim_{x \rightarrow 0}\frac{(x^2 -1)6x}{x^3 +3x^2 +2} \\
&= \lim_{x \rightarrow 0} \frac{6x^3-6x}{x^3 +3x^2 +2} \\
&= \frac{18x^2-6}{x^3 +3x^2 +2} \\
&= \frac{-6}{2} \\
& = -3
\end{align*}
Jawaban : A
Soal Nomor 8. \( \lim_{x \rightarrow 1}\frac{(x^2 +x - 2)\sin (x-1)}{x^2 -2x +1} = \) ....
A. 4
B. 3
C. 0
D. \(-\frac{1}{4} \)
E. \(-\frac{1}{2} \)
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 1}\frac{(x^2 +x - 2)\sin (x-1)}{x^2 -2x +1} &= \lim_{x \rightarrow 1}\frac{(x-1)(x+2)(x-1)}{(x-1)^2} \\
&= \lim_{x \rightarrow 1} \frac{6x^3-6x}{x^3 +3x^2 +2} \\
&= \frac{x+2}{1} \\
&= \frac{3}{1} \\
& = 3
\end{align*}
Jawaban : B
Soal Nomor 9. \( \lim_{t \rightarrow 2}\frac{(t-2)(t-3)\sin (t-2)}{[(t-2)(t+1)]^2} = \) ....
A. \(\frac{1}{3} \)
B. \(\frac{1}{9} \)
C. 0
D. \(-\frac{1}{9} \)
E. \(-\frac{1}{3} \)
Pembahasan :
\begin{align*}
\lim_{t \rightarrow 2}\frac{(t-2)(t-3)\sin (t-2)}{[(t-2)(t+1)]^2} &= \lim_{x \rightarrow 2}\frac{(t-2)(t-3)(t-2)}{[(t-2)(t+1)]^2} \\
&= \lim_{t \rightarrow 2} \frac{(t-3)}{(t+1)^2} \\
&= \lim_{t \rightarrow 2} \frac{x+2}{1} \\
&= \frac{1}{9}
\end{align*}
Jawaban : B
Soal Nomor 10. \( \lim_{x \rightarrow \pi}\frac{x-\pi}{2(x-\pi) + \tan (x - \pi)} = \) ....
A. \(-\frac{1}{2} \)
B. \(-\frac{1}{4} \)
C. \(\frac{1}{4} \)
D. \(\frac{1}{3} \)
E. \(\frac{2}{5} \)
Pembahasan :
\begin{align*}
\lim_{x \rightarrow \pi}\frac{x-\pi}{2(x-\pi) + \tan (x - \pi)} &= \lim_{x \rightarrow \pi}\frac{x-\pi}{2(x-\pi)+(x-\pi)} \\
&= \lim_{x \rightarrow \pi} \frac{(x-\pi)}{3(x-\pi)} \\
&= \frac{1}{3}
\end{align*}
Jawaban : D
--oo0oo--
A. -1
B. 0
C. 1
D. 2
E. 3
Pembahasan :
Cara 1 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{9x - \tan x}{x + \sin 3x} &= \lim_{x \rightarrow 0}\frac{9 -\frac{1}{x}\tan x}{1 + \frac{1}{x} \sin 3x} \\
&= \frac{9-1}{1+3} \\
&= 2
\end{align*}
Cara 2 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{9x - \tan x}{x + \sin 3x} &= \lim_{x \rightarrow 0}\frac{9x -x}{x + 3x} \\
&= \frac{8x}{4x} \\
&= 2
\end{align*}
Jawaban : D
Soal Nomor 2. \( \lim_{x \rightarrow 0}\frac{3x - \sin 4x}{5x + \tan 2x} = \) ....
A. \(\frac{3}{5} \)
B. -1
C. 2
D. \(\frac{7}{3} \)
E. ∞
Pembahasan :
Cara 1 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{3x - \sin 4x}{5x + \tan 2x} &= \lim_{x \rightarrow 0}\frac{3 +\frac{1}{x}\sin 4x}{5x - \frac{1}{x}\tan 2x} \\
&= \frac{3+4}{5-2} \\
&= \frac{7}{3}
\end{align*}
Cara 2 :
\begin{align*}
\lim_{x \rightarrow 0}\frac{3x - \sin 4x}{5x + \tan 2x} &= \lim_{x \rightarrow 0}\frac{3x +4x}{5x - 2x} \\
&= \lim_{x \rightarrow 0} \frac{7x}{3x} \\
&= \frac{7}{3}
\end{align*}
Jawaban : D
Soal Nomor 3. \( \lim_{x \rightarrow k}\frac{2k - 2x}{\sin (x-k)+ \tan (3x-3k)} = \) ....
A. -1
B. \(-\frac{1}{2} \)
C. 0
D. \(\frac{1}{2} \)
E. 1
Pembahasan :
\begin{align*}
\lim_{x \rightarrow k}\frac{2k - 2x}{\sin (x-k)+ \tan (3x-3k)} &= \lim_{x \rightarrow k}\frac{2k - 2x}{(x-k)+(3x - 3k)} \\
&= \lim_{x \rightarrow k} \frac{-2(x-k)}{4(x-k)} \\
&= \frac{-2}{4} \\
&= -\frac{1}{2}
\end{align*}
Jawaban : B
Soal Nomor 4. \( \lim_{x \rightarrow a}\frac{7\sin (x-a) +5\tan (x-a)}{x-a + \sin (x-a)} = \) ....
A. 0
B. 2
C. 4
D. 6
E. 8
Pembahasan :
\begin{align*}
\lim_{x \rightarrow a}\frac{7\sin (x-a) +5\tan (x-a)}{x-a + \sin (x-a)} &= \lim_{x \rightarrow a}\frac{7(x-a) +5(x-a)}{x-a + (x-a)} \\
&= \lim_{x \rightarrow a} \frac{12(x-a)}{2(x-a)} \\
&= \frac{12}{2} \\
&= 6
\end{align*}
Jawaban : D
Soal Nomor 5. \( \lim_{x \rightarrow 0}\frac{\sin 4x \tan ^2 3x +6x^3}{2x^2 \sin 3x \cos 2x} = \) ....
A. 0
B. 3
C. 4
D. 5
E. 7
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 0}\frac{\sin 4x \tan ^2 3x +6x^3}{2x^2 \sin 3x \cos 2x} &= \lim_{x \rightarrow 0}\frac{4x (3x)^2 +6x^3}{2x^2 (3x) 1} \\
&= \lim_{x \rightarrow 0} \frac{42x^3}{6x^3} \\
&= \frac{42}{6} \\
&= 7
\end{align*}
Jawaban : E
Soal Nomor 6. \( \lim_{x \rightarrow 0}\frac{\sin (2x^2)}{x^2 +(\sin 3x)^2} = \) ....
A. \(\frac{2}{3} \)
B. 5
C. \(\frac{3}{2} \)
D. 0
E. \(\frac{1}{5} \)
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 0}\frac{\sin (2x^2)}{x^2 +(\sin 3x)^2} &= \lim_{x \rightarrow 0}\frac{2x^2}{x^2 + (3x)^2} \\
&= \lim_{x \rightarrow 0} \frac{3x^2}{10x^2} \\
&= \frac{2}{10} \\
&= \frac{1}{5}
\end{align*}
Jawaban : E
Soal Nomor 7. \( \lim_{x \rightarrow 0}\frac{(x^2 -1)\sin 6x}{x^3 +3x^2 +2} = \) ....
A. -3
B. -1
C. 0
D. 1
E. 6
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 0}\frac{(x^2 -1)\sin 6x}{x^3 +3x^2 +2} &= \lim_{x \rightarrow 0}\frac{(x^2 -1)6x}{x^3 +3x^2 +2} \\
&= \lim_{x \rightarrow 0} \frac{6x^3-6x}{x^3 +3x^2 +2} \\
&= \frac{18x^2-6}{x^3 +3x^2 +2} \\
&= \frac{-6}{2} \\
& = -3
\end{align*}
Jawaban : A
Soal Nomor 8. \( \lim_{x \rightarrow 1}\frac{(x^2 +x - 2)\sin (x-1)}{x^2 -2x +1} = \) ....
A. 4
B. 3
C. 0
D. \(-\frac{1}{4} \)
E. \(-\frac{1}{2} \)
Pembahasan :
\begin{align*}
\lim_{x \rightarrow 1}\frac{(x^2 +x - 2)\sin (x-1)}{x^2 -2x +1} &= \lim_{x \rightarrow 1}\frac{(x-1)(x+2)(x-1)}{(x-1)^2} \\
&= \lim_{x \rightarrow 1} \frac{6x^3-6x}{x^3 +3x^2 +2} \\
&= \frac{x+2}{1} \\
&= \frac{3}{1} \\
& = 3
\end{align*}
Jawaban : B
Soal Nomor 9. \( \lim_{t \rightarrow 2}\frac{(t-2)(t-3)\sin (t-2)}{[(t-2)(t+1)]^2} = \) ....
A. \(\frac{1}{3} \)
B. \(\frac{1}{9} \)
C. 0
D. \(-\frac{1}{9} \)
E. \(-\frac{1}{3} \)
Pembahasan :
\begin{align*}
\lim_{t \rightarrow 2}\frac{(t-2)(t-3)\sin (t-2)}{[(t-2)(t+1)]^2} &= \lim_{x \rightarrow 2}\frac{(t-2)(t-3)(t-2)}{[(t-2)(t+1)]^2} \\
&= \lim_{t \rightarrow 2} \frac{(t-3)}{(t+1)^2} \\
&= \lim_{t \rightarrow 2} \frac{x+2}{1} \\
&= \frac{1}{9}
\end{align*}
Jawaban : B
Soal Nomor 10. \( \lim_{x \rightarrow \pi}\frac{x-\pi}{2(x-\pi) + \tan (x - \pi)} = \) ....
A. \(-\frac{1}{2} \)
B. \(-\frac{1}{4} \)
C. \(\frac{1}{4} \)
D. \(\frac{1}{3} \)
E. \(\frac{2}{5} \)
Pembahasan :
\begin{align*}
\lim_{x \rightarrow \pi}\frac{x-\pi}{2(x-\pi) + \tan (x - \pi)} &= \lim_{x \rightarrow \pi}\frac{x-\pi}{2(x-\pi)+(x-\pi)} \\
&= \lim_{x \rightarrow \pi} \frac{(x-\pi)}{3(x-\pi)} \\
&= \frac{1}{3}
\end{align*}
Jawaban : D
--oo0oo--
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