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Matematika Dasar Limit Fungsi Trigonometri (*Soal Dan Pembahasan)

Rabu, 11 Januari 2017 Add Comment Edit

Limit Kiri = Limit Kanan=L

$\underset{x \to 0}{lim} \dfrac{sin\ x }{x} = 1 , \, \, $ atau $\underset{x \to 0}{lim}\dfrac{ x }{sin\ x} = 1$

$\underset{x \to 0}{lim} \dfrac{tan\ x }{x} = 1 , \, \, $ atau $\underset{x \to 0}{lim}\dfrac{ x }{tan\ x} = 1$

$\underset{x \to 0}{lim} \dfrac{sin\ ax }{bx} = \dfrac{a}{b} , \, \, $ atau $\underset{x \to 0}{lim}\dfrac{ ax }{sin\ bx} = \dfrac{a}{b}$

$\underset{x \to 0}{lim} \dfrac{tan\ ax }{bx} = \dfrac{a}{b} , \, \, $ atau $\underset{x \to 0}{lim}\dfrac{ ax }{tan\ bx} = \dfrac{a}{b}$

$\underset{x \to 0}{lim} \dfrac{sin\ ax }{sin\ bx} = \dfrac{a}{b} , \, \, $ atau $\underset{x \to 0}{lim}\dfrac{tan\ ax }{tan\ bx} = \dfrac{a}{b}$

$\underset{x \to 0}{lim} \dfrac{tan\ ax }{sin\ bx} = \dfrac{a}{b} , \, \, $ atau $\underset{x \to 0}{lim}\dfrac{sin\ ax }{tan\ bx} = \dfrac{a}{b}$

Identitas Trigonometri Dasarteorema limit pada fungsi aljabarMetode L'Hospital (Turunan)1. Soal STIS 2017 (*Soal Lengkap)$\underset{x \to 2}{lim} \dfrac{\left( x^{2}-5x-6\right)\ sin\ 2(x-2) }{\left( x^{2}-x-2\right)} \cdots$
$\begin{align}
(A)\ & -8 \\
(B)\ & -5 \\
(C)\ & -2 \\
(D)\ & \dfrac{3}{4} \\
(E)\ & 5
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \underset{x \to 2}{lim} \dfrac{\left( x^{2}-5x-6\right)\ sin\ 2(x-2) }{\left( x^{2}-x-2 \right)} \\
& = \underset{x \to 2}{lim} \dfrac{(x+1)(x-6)\ sin\ 2(x-2) }{(x-2)(x+1)} \\
& = \underset{x \to 2}{lim} \dfrac{(x-6)\ sin\ 2(x-2) }{(x-2)} \\
& = \underset{x \to 2}{lim} (x-6) \cdot \underset{x \to 2}{lim} \dfrac{sin\ 2(x-2) }{(x-2)} \\
& = (2-6) \cdot 2 \\
& = -4 \cdot 2 =-8 \\
\end{align}$

$\therefore$ Pilihan yang sesuai $(A)\ - 8$

2. Soal SBMPTN 2018 (*Soal Lengkap)$\underset{x \to 2}{lim} \dfrac{sin\ \left( 2x-4 \right) }{2- \sqrt{6-x}} =\cdots$
$\begin{align}
(A)\ & -8 \\
(B)\ & -2 \\
(C)\ & 0 \\
(D)\ & 2 \\
(E)\ & 8
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \underset{x \to 2}{lim} \dfrac{sin\ \left( 2x-4 \right) }{2- \sqrt{6-x}} \\
& = \underset{x \to 2}{lim} \dfrac{sin\ \left( 2x-4 \right) }{2- \sqrt{6-x}} \cdot \dfrac{2+ \sqrt{6-x}}{2+ \sqrt{6-x}} \\
& = \underset{x \to 2}{lim} \dfrac{sin\ \left( 2x-4 \right) \left( 2+ \sqrt{6-x} \right) }{4- \left( 6-x \right)} \\
& = \underset{x \to 2}{lim} \dfrac{sin\ \left( 2x-4 \right) \left( 2+ \sqrt{6-x} \right) }{4- 6+x } \\
& = \underset{x \to 2}{lim} \dfrac{sin\ 2\left( x-2 \right) \left( 2+ \sqrt{6-x} \right) }{x-2 } \\
& = \underset{x \to 2}{lim} \dfrac{sin\ 2\left( x-2 \right)}{x-2 } \cdot \underset{x \to 2}{lim} \left( 2+ \sqrt{6-x} \right) \\
& = 2 \cdot \left( 2+ \sqrt{6-2} \right) \\
& = 2 \cdot ( 2+ 2)=8 \\
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ 8$

3. Soal SBMPTN 2017 (*Soal Lengkap)$\underset{x \to 0}{lim} \dfrac{sec\ x+cos\ x-2}{x^{2}\ sin^{2}x}=\cdots$
$\begin{align}
(A)\ & -\dfrac{1}{8} \\
(B)\ & -\dfrac{1}{4} \\
(C)\ & 0 \\
(D)\ & \dfrac{1}{4} \\
(E)\ & \dfrac{1}{8}
\end{align}$Alternatif Pembahasan: show

Identitas trigonometri yg mungkin diperlukan:
$cos\ 4x=cos^{2}2x-sin^{2}2x$
$cos\ 2x=cos^{2}x-sin^{2}x$
$cos\ x=cos^{2} \dfrac{1}{2}x-sin^{2}\dfrac{1}{2}x$
$1=cos^{2} \dfrac{1}{2}x+sin^{2}\dfrac{1}{2}x$
$cos\ x - 1=-2sin^{2}\dfrac{1}{2}x$

Kita kembali ke soal;
$\begin{align}
& \underset{x \to 0}{lim} \dfrac{sec\ x+cos\ x-2}{x^{2}\ sin^{2}x}\\
= & \underset{x \to 0}{lim} \dfrac{\dfrac{1}{cos\ x}+\dfrac{cos^{2}x}{cos\ x}-\dfrac{2\ cos\ x}{cos\ x}}{x^{2}\ sin^{2}x}\\
= & \underset{x \to 0}{lim} \dfrac{cos^{2}-2\ cos\ x+1}{x^{2}\ sin^{2}x\ cos\ x}\\
= & \underset{x \to 0}{lim} \dfrac{\left (cos\ x-1 \right )^{2}}{x^{2}\ sin^{2}x\ cos\ x}\\
= & \underset{x \to 0}{lim} \dfrac{\left (-2sin^{2}(\dfrac{1}{2}x) \right )^{2}}{x^{2}\ sin^{2}x\ cos\ x}\\
= & \underset{x \to 0}{lim} \dfrac{4\ sin^{2}(\dfrac{1}{2}x)\ sin^{2}(\dfrac{1}{2}x)}{x^{2}\ sin^{2}x\ cos\ x}\\
= & \underset{x \to 0}{lim} 4\ \cdot \dfrac{sin^{2}(\dfrac{1}{2}x)}{x^{2}} \cdot \dfrac{sin^{2}(\dfrac{1}{2}x)}{sin^{2}x} \cdot \dfrac{1}{cos\ x}\\
= & 4\ \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{1} = \dfrac{1}{4}
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ \dfrac{1}{4}$

4. Soal SBMPTN 2016 (*Soal Lengkap)$\underset{x \to 0}{lim} \dfrac{x^{3}}{\sqrt{1+sin\ x}-\sqrt{1+tan\ x}}=\cdots$
$\begin{align}
(A)\ & -4 \\
(B)\ & -2 \\
(C)\ & 0 \\
(D)\ & 2 \\
(E)\ & 4
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \underset{x \to 0}{lim} \dfrac{x^{3}}{\sqrt{1+sin\ x}-\sqrt{1+tan\ x}} \\
& = \underset{x \to 0}{lim} \dfrac{x^{3}}{\sqrt{1+sin\ x}-\sqrt{1+tan\ x}} \cdot \dfrac{\sqrt{1+sin\ x}+\sqrt{1+tan\ x}}{\sqrt{1+sin\ x}+\sqrt{1+tan\ x}} \\
& \underset{x \to 0}{lim} \dfrac{x^{3} \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right)}{1+sin\ x-1-tan\ x} \\
& = \underset{x \to 0}{lim} \dfrac{x^{3} \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right)}{sin\ x-tan\ x} \\
& = \underset{x \to 0}{lim} \dfrac{x^{3} \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right)}{sin\ x (1-\frac{1}{cos\ x})} \\
& = \underset{x \to 0}{lim} \dfrac{x^{3} \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right)}{sin\ x \cdot \frac{cos\ x -1}{cos\ x}} \\
& = \underset{x \to 0}{lim} \dfrac{cos\ x \cdot x^{3} \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right)}{sin\ x (cos\ x -1)} \\
& = \underset{x \to 0}{lim} \dfrac{cos\ x \cdot x^{3} \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right)}{sin\ x (1-2 sin^{2} \frac{1}{2}x -1)} \\
& = \underset{x \to 0}{lim} \dfrac{cos\ x \cdot x^{3} \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right)}{sin\ x (-2 sin^{2} \frac{1}{2}x)} \\
& = \underset{x \to 0}{lim}\ cos\ x \left( \sqrt{1+sin\ x}+\sqrt{1+tan\ x} \right) \cdot \underset{x \to 0}{lim} \dfrac{x^{3}}{sin\ x (-2 sin^{2} \frac{1}{2}x)}\\
& = cos\ 0 \left( \sqrt{1+sin\ 0}+\sqrt{1+tan\ 0} \right) \cdot \dfrac{1}{-2 \cdot \frac{1}{2} \cdot \frac{1}{2}} \\
& = 1 \left( \sqrt{1}+\sqrt{1} \right) \cdot \dfrac{1}{-\frac{1}{2}}\\
& = 2 \cdot (-2) =-4
\end{align}$

$\therefore$ Pilihan yang sesuai $(A)\ -4$

5. Soal SBMPTN 2013 (*Soal Lengkap)$\underset{x \to 0}{lim} \dfrac{x\ tan\ x}{x\ sin\ x - cos\ x +1}=\cdots$
$\begin{align}
(A)\ & 2 \\
(B)\ & \frac{3}{2} \\
(C)\ & 1 \\
(D)\ & \frac{2}{3} \\
(E)\ & -1
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{x\ sin\ x - cos\ x +1} \\
&= \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{x\ sin\ x +1- cos\ x} \\
&= \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{x\ sin\ x +2 sin^{2} \dfrac{1}{2}x} \\
&= \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{x\ sin\ x +2 sin^{2} \dfrac{1}{2}x} \cdot \dfrac{\dfrac{1}{x^{2}}}{\dfrac{1}{x^{2}}} \\
&= \underset{x \to 0}{lim} \dfrac{\dfrac{x\ tan\ x}{x^{2}}}{\dfrac{x\ sin\ x}{x^{2}} +\dfrac{2 sin^{2} \dfrac{1}{2}x}{x^{2}}} \\
&= \dfrac{1}{1+2 \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}} \\
&= \dfrac{1}{1+\dfrac{1}{2}} \\
&= \dfrac{1}{\dfrac{3}{2}}= \dfrac{2}{3}
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ \dfrac{2}{3}$

6. Soal SBMPTN 2013 (*Soal Lengkap)$\underset{x \to 0}{lim} \sqrt{\dfrac{x\ tan\ x}{sin^{2} x - cos\ 2x +1}}=\cdots$
$\begin{align}
(A)\ & 3 \\
(B)\ & \sqrt{3} \\
(C)\ & \frac{\sqrt{3}}{3} \\
(D)\ & \frac{1}{3} \\
(E)\ & \frac{\sqrt{3}}{2}
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \underset{x \to 0}{lim} \sqrt{\dfrac{x\ tan\ x}{sin^{2} x - cos\ 2x +1}} \\
&= \underset{x \to 0}{lim} \sqrt{\dfrac{x\ tan\ x}{sin^{2} x + 1- cos\ 2x }} \\
&= \underset{x \to 0}{lim} \sqrt{\dfrac{x\ tan\ x}{sin^{2} x + 2sin^{2} x }} \\
&= \underset{x \to 0}{lim} \sqrt{\dfrac{x\ tan\ x}{3 sin^{2} x }} \\
&= \underset{x \to 0}{lim} \sqrt{\dfrac{x\ tan\ x}{3 sin\ x\ sin\ x}} \\
&= \underset{x \to 0}{lim} \sqrt{\dfrac{1}{3} \cdot \dfrac{x}{sin\ x} \cdot \dfrac{tan\ x}{sin\ x}} \\
&= \sqrt{\dfrac{1}{3} \cdot 1 \cdot 1} \\
&= \sqrt{\dfrac{1}{3}} \\
&= \dfrac{1}{3}\sqrt{3}
\end{align}$

$\therefore$ Pilihan yang sesuai $(C)\ \frac{\sqrt{3}}{3}$

7. Soal SBMPTN 2013 (*Soal Lengkap)$\underset{x \to 0}{lim} \dfrac{x\ tan\ x}{sin^{2} x - cos\ 2x +1}=\cdots$
$\begin{align}
(A)\ & 1 \\
(B)\ & \frac{1}{3} \\
(C)\ & \frac{2}{3} \\
(D)\ & -\frac{1}{2} \\
(E)\ & -1
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{sin^{2} x - cos\ 2x +1} \\
&= \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{sin^{2} x + 1- cos\ 2x } \\
&= \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{sin^{2} x + 2sin^{2} x } \\
&= \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{3 sin^{2} x } \\
&= \underset{x \to 0}{lim} \dfrac{x\ tan\ x}{3 sin\ x\ sin\ x} \\
&= \underset{x \to 0}{lim} \dfrac{1}{3} \cdot \dfrac{x}{sin\ x} \cdot \dfrac{tan\ x}{sin\ x} \\
&= \dfrac{1}{3} \cdot 1 \cdot 1 \\
&= \dfrac{1}{3}
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ \frac{1}{3}$

8. Soal UM UGM 2017 $\underset{x \to -4}{lim} \dfrac{1-cos(x+4)}{x^{2}+8x+16}=\cdots$
$\begin{align}
(A)\ & -2 \\
(B)\ & -\frac{1}{2} \\
(C)\ & \frac{1}{3} \\
(D)\ & \frac{1}{2} \\
(E)\ & 2
\end{align}$Alternatif Pembahasan: show

$\begin{align}
& \underset{x \to -4}{lim} \dfrac{1-cos(x+4)}{x^{2}+8x+16} \\
& = \underset{x \to -4}{lim} \dfrac{1-cos(x+4)}{(x+4)(x+4)} \\
& = \underset{x \to -4}{lim} \dfrac{2 sin^{2} \dfrac{1}{2}(x+4)}{(x+4)(x+4)} \\
& = \underset{x \to -4}{lim} \dfrac{2 sin \dfrac{1}{2}(x+4)}{(x+4)} \cdot \underset{x \to -4}{lim} \dfrac{sin \dfrac{1}{2}(x+4)}{(x+4)} \\
& = 2 \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \\
& = \dfrac{1}{2}
\end{align}$

$\therefore$ Pilihan yang sesuai $(D)\ \frac{1}{2}$

9. Soal UM UNDIP 2010 (*Soal Lengkap)$\underset{x \to y}{lim} \dfrac{sin\ x - sin\ y}{x-y}=\cdots$
$\begin{align}
(A)\ & sin\ x \\
(B)\ & sin\ y \\
(C)\ & 0 \\
(D)\ & cos\ x \\
(E)\ & cos\ y
\end{align}$Alternatif Pembahasan: show

Untuk menuntaskan bentuk ini, kita gunakan sedikit identitas trigonometri yaitu $sin\ x - sin\ y$ ialah $2\ cos\ \dfrac{1}{2}(x+y)\ sin\ \dfrac{1}{2}(x-y)$.

$\begin{align}
& \underset{x \to y}{lim} \dfrac{sin\ x - sin\ y}{x-y} \\
& = \underset{x \to y}{lim} \dfrac{2\ cos\ \dfrac{1}{2}(x+y)\ sin\ \dfrac{1}{2}(x-y)}{x-y} \\
& = \underset{x \to y}{lim}\ 2\ cos\ \dfrac{1}{2}(x+y) \times \underset{x \to y}{lim} \dfrac{sin\ \dfrac{1}{2}(x-y)}{x-y} \\
& = 2\ cos\ \dfrac{1}{2}(y+y) \times \dfrac{1}{2} \\
& = cos\ \dfrac{1}{2}(2y) \\
& = cos\ y
\end{align}$

$\therefore$ Pilihan yang sesuai $(E)\ cos\ y$

10. Soal UM UNDIP 2010 (*Soal Lengkap)$\underset{x \to -1}{lim} \dfrac{sin(1-x^{2})\ cos (1-x^{2})}{x^{2}-1}=\cdots$
$\begin{align}
(A)\ & 1 \\
(B)\ & -1 \\
(C)\ & 2 \\
(D)\ & -2 \\
(E)\ & 0
\end{align}$Alternatif Pembahasan: show

Untuk menuntaskan soal diatas kita coba dengan memisalkan $1-x^{2}=m$, alasannya ialah $x \to -1$ maka $m \to 0$.
$\begin{align}
& \underset{x \to -1}{lim} \dfrac{sin(1-x^{2})\ cos (1-x^{2})}{x^{2}-1} \\
& = \underset{m \to 0}{lim} \dfrac{sin\ m\ cos\ m}{-m} \\
& = \underset{m \to 0}{lim}\ cos\ m \cdot \underset{m \to 0}{lim} \dfrac{sin\ m}{-m} \\
& = cos\ 0 \cdot -1 \\
& = 1 \cdot -1 =-1
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ -1$

11. Soal Latihan Matematika$\underset{x \to 0}{lim} \dfrac{2x+sin\ 5x}{4x+tan\ 7x}=\cdots$
$\begin{align}
(A)\ & \dfrac{5}{14} \\
(B)\ & \dfrac{1}{2} \\
(C)\ & \dfrac{7}{11} \\
(D)\ & \dfrac{5}{7} \\
(E)\ & \dfrac{11}{7} \\
\end{align}$Alternatif Pembahasan: show

Untuk menuntaskan soal diatas kita coba dengan memanipulasi aljabar dengan mengalikan dengan $satu$
$\begin{align}
& \underset{x \to 0}{lim} \dfrac{2x+sin\ 5x}{4x+tan\ 7x}\\
& = \underset{x \to 0}{lim} \dfrac{2x+sin\ 5x}{4x+tan\ 7x} \times \dfrac{\dfrac{1}{sin\ x}}{\dfrac{1}{sin\ x}} \\
& = \underset{x \to 0}{lim} \dfrac{\dfrac{2x+sin\ 5x}{sin\ x}}{\dfrac{4x+tan\ 7x}{sin\ x}} \\
& = \underset{x \to 0}{lim} \dfrac{\dfrac{2x}{sin\ x}+\dfrac{sin\ 5x}{sin\ x}}{\dfrac{4x}{sin\ x}+\dfrac{tan\ 7x}{sin\ x}} \\
& = \dfrac{2+5}{4+7}\\
& = \dfrac{7}{11}
\end{align}$

$\therefore$ Pilihan yang sesuai $(C)\ \dfrac{7}{11}$

12. Soal Latihan Matematika$\underset{x \to \dfrac{\pi}{4}}{lim} \dfrac{sin\ \dfrac{\pi}{4} - sin\ x}{1-x}=\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & 0 \\
(C)\ & 1 \\
(D)\ & \dfrac{1}{2} \sqrt{2} \\
(E)\ & \sqrt{2}
\end{align}$Alternatif Pembahasan: show

Untuk menuntaskan soal diatas kita sanggup eksklusif dengan mensubstitusi langsung, yaitu:
$\begin{align}
& \underset{x \to \dfrac{\pi}{4}}{lim} \dfrac{sin\ \dfrac{\pi}{4} - sin\ x}{1-x} \\
& = \dfrac{sin\ \dfrac{\pi}{4} - sin\ \dfrac{\pi}{4}}{1-\dfrac{\pi}{4}} \\
& = \dfrac{0}{1-\dfrac{\pi}{4}} \\
& = 0
\end{align}$

$\therefore$ Pilihan yang sesuai $(B)\ 0$

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Matematika Dasar Limit Fungsi Trigonometri Matematika Dasar Limit Fungsi Trigonometri (*Soal dan Pembahasan)

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Matematika Dasar Limit Fungsi Trigonometri (*Soal Dan Pembahasan)