SMA pilpay: Matematika Dasar Limit Tak Sampai (*Soal Dari Banyak Sekali Sumber)

Friday, November 30, 2018

Matematika Dasar Limit Tak Sampai (*Soal Dari Banyak Sekali Sumber)

"Sir, i have a problem with limit function from SIMAK UI 2009"Nilai $ \underset{x \to \infty}{lim} \left ( \sqrt{4x^2+8x}-\sqrt{x^2+1}-\sqrt{x^2+x} \right ) \cdots$
$\begin{align}
(A)\ & \frac{5}{2} \\
(B)\ & 2 \\
(C)\ & \frac{3}{2} \\
(D)\ & 1 \\
(E)\ & \frac{1}{2}
\end{align}$

$\underset{x \to 0}{lim} \dfrac{a}{x^{n}}=\text{tidak mempunyai nilai limit}$; untuk $n$ bilangan orisinil ganjil

$\underset{x \to 0}{lim} \dfrac{a}{x^{n}}= \infty $; untuk $n$ bilangan orisinil genap

$\underset{x \to \infty}{lim}\ ax^{n}= \infty $; untuk $n$ bilangan asli

$\underset{x \to \infty}{lim} \dfrac{k}{x^{n}}= 0 $; untuk $n$ bilangan asli

$\underset{x \to \infty}{lim} \dfrac{ax^m + a_1x^{m-1} + ...}{bx^n + b_1 x^{n-1} + ....}=\infty $; untuk $m \gt n$ bilangan asli

$\underset{x \to \infty}{lim} \dfrac{ax^m + a_1x^{m-1} + ...}{bx^n + b_1 x^{n-1} + ....}=\dfrac{a}{b}$; untuk $m=n$ bilangan asli

$\underset{x \to \infty}{lim} \dfrac{ax^m + a_1x^{m-1} + ...}{bx^n + b_1 x^{n-1} + ....}=0$; untuk $m \lt n$ bilangan asli

$\underset{x \to \infty}{lim} \sqrt{ax^{2} + bx + c } - \sqrt{ax^{2} + qx + r } = \frac{b-q}{2\sqrt{a}}$

$\underset{x \to \infty}{lim} \sqrt{ax^{2} + bx + c } - \sqrt{px^{2} + qx + r } = + \infty $; untuk $a \gt p$

$\underset{x \to \infty}{lim} \sqrt{ax^{2} + bx + c } - \sqrt{px^{2} + qx + r } = - \infty $; untuk $a \lt p$

$\underset{x \to \infty}{lim} \sqrt[n]{ax^{n} + bx^{n-1} + \cdots } - \sqrt[n]{ax^{n} + qx^{n-1} + \cdots } = \frac{b-q}{ n \cdot \sqrt[n]{a^{n-1}}}$

$\underset{x \to \infty}{lim} \sqrt[n]{ax^{n} + bx^{n-1} + \cdots } - \sqrt[n]{px^{n} + qx^{n-1} + \cdots } = \infty$; untuk $a \gt p$

$\underset{x \to \infty}{lim} \sqrt[n]{ax^{n} + bx^{n-1} + \cdots } - \sqrt[n]{px^{n} + qx^{n-1} + \cdots } = 0$; untuk $a \lt p$

1. Soal SIMAK UI 2009 [Soal Lengkap Disini]Nilai $ \underset{x \to \infty}{lim} \left ( \sqrt{4x^2+8x}-\sqrt{x^2+1}-\sqrt{x^2+x} \right ) \cdots$
$\begin{align}
(A)\ & \frac{5}{2} \\
(B)\ & 2 \\
(C)\ & \frac{3}{2} \\
(D)\ & 1 \\
(E)\ & \frac{1}{2}
\end{align}$Alternatif Pembahasan:

Hint

Penyelesaian soal coba dianalisis oleh Heryanto Simatupang;
$\begin{align}
& \underset{x \to \infty}{lim} \left ( \sqrt{4x^2+8x}-\sqrt{x^2+1}-\sqrt{x^2+x} \right ) \\
& = \underset{x \to \infty}{lim} \left ( \sqrt{4x^2+8x}-\sqrt{x^2+1}-\sqrt{x^2+x} \right ) \\
& = \underset{x \to \infty}{lim} \left ( \sqrt{4\left (x^2+2x \right )}-\sqrt{x^2+1}-\sqrt{x^2+x} \right ) \\
& = \underset{x \to \infty}{lim} \left ( 2 \sqrt{\left (x^2+2x \right )}-\sqrt{x^2+1}-\sqrt{x^2+x} \right ) \\
& = \underset{x \to \infty}{lim} \left (\sqrt{x^2+2x}+ \sqrt{x^2+2x }-\sqrt{x^2+1}-\sqrt{x^2+x} \right ) \\
& = \underset{x \to \infty}{lim} \left (\sqrt{x^2+2x}- \sqrt{x^2+1 } +\sqrt{x^2+2x}-\sqrt{x^2+x} \right ) \\
& = \underset{x \to \infty}{lim} \left (\sqrt{x^2+2x}- \sqrt{x^2+1 }\right ) +\lim_{x\rightarrow \infty }\left (\sqrt{x^2+2x}-\sqrt{x^2+x} \right ) \\
& = \frac{2-0}{2\sqrt{1}}+\frac{2-1}{2\sqrt{1}} \\
& = 1+ \frac{1}{2} \\
& = \frac{3}{2}
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(C)\ \frac{3}{2}$

2. Soal UM UGM 2003Nilai dari limit fungsi $ \underset{x \to \infty}{lim} \left ( \sqrt{2x^2+5x+6}-\sqrt{2x^2+2x-1}\right )=\cdots$
$\begin{align}
(A)\ & \frac{3}{2}\sqrt{2} \\
(B)\ & \frac{3}{4}\sqrt{2} \\
(C)\ & - \frac{3}{2}\sqrt{2} \\
(D)\ & - \frac{3}{4}\sqrt{2} \\
(E)\ & 3
\end{align}$Alternatif Pembahasan:

Hint

$\begin{align}
& \underset{x \to \infty}{lim} \left ( \sqrt{2x^2+5x+6}-\sqrt{2x^2+2x-1}\right ) \\
& = \frac{b-q}{2\sqrt{a}} \\
& = \frac{5-2}{2\sqrt{2}} \\
& = \frac{3}{2\sqrt{2}} \\
& = \frac{3\sqrt{2}}{4} \\
& = \frac{3}{4}\sqrt{2}
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(B)\ \frac{3}{4}\sqrt{2}$

3. Soal UN Sekolah Menengan Atas 2016 [Soal Lengkap Disini]Nilai dari limit fungsi $\underset{x \to \infty}{lim} \left ( \sqrt{4x^2+4x-3}- \left (2x-5 \right ) \right )=\cdots$
$\begin{align}
(A)\ & -6 \\
(B)\ & -4 \\
(C)\ & -1 \\
(D)\ & 4 \\
(E)\ & 6
\end{align}$Alternatif Pembahasan:

Hint

$\begin{align}
& \underset{x \to \infty}{lim} \left ( \sqrt{4x^2+4x-3}- \left (2x-5 \right ) \right ) \\
& =\underset{x \to \infty}{lim} \left ( \sqrt{4x^2+4x-3}- \left (2x-5 \right ) \right ) \\
& =\underset{x \to \infty}{lim} \left ( \sqrt{4x^2+4x-3}-\sqrt{ \left (2x-5 \right )^{2}} \right ) \\
& =\underset{x \to \infty}{lim} \left ( \sqrt{4x^2+4x-3}-\sqrt{4x^2-20x+25} \right ) \\
& =\frac{b-q}{2\sqrt{a}} \\
& =\frac{4+20}{2\sqrt{4}} \\
& =\frac{24}{4} \\
& =6
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(E)\ 6$

4. Soal SIMAK UI 2010 [Soal Lengkap Disini]$ \underset{x \to \infty}{lim} \left ( \sqrt{64x^2+ax+7}-8x+b \right )= \frac{3}{2} $.
Jika $a$ dan $b$ bilangan lingkaran positif, maka nilai $a+b$ adalah...
$\begin{align}
(A)\ & 5 \\
(B)\ & 9 \\
(C)\ & 12 \\
(D)\ & 16 \\
(E)\ & 24
\end{align}$Alternatif Pembahasan:

Hint

$\begin{align}
\frac{3}{2} & = \underset{x \to \infty}{lim} \left ( \sqrt{64x^2+ax+7}-8x+b \right ) \\
& =\underset{x \to \infty}{lim} \left ( \sqrt{64x^2+ax+7}-\left (8x-b \right ) \right ) \\
& =\underset{x \to \infty}{lim} \left ( \sqrt{64x^2+ax+7}-\sqrt{\left (8x-b \right )^{2}}\right ) \\
& =\underset{x \to \infty}{lim} \left ( \sqrt{64x^2+ax+7}-\sqrt{64x^2-16bx+b^{2}}\right ) \\
& =\frac{a+16b}{2\sqrt{64}} \\
\frac{3}{2} & =\frac{a+16b}{16} \\
24 & = a+16b
\end{align}$

Karena $a$ dan $b$ ialah bilangan lingkaran faktual maka nilai $a$ dan $b$ yang memenuhi persamaan $a+16b= 24$ ialah ketika $b=1$ dan $a=8$, maka $a+b=9$

$\therefore$ Pilihan yang sesuai ialah $(B)\ 9$

5. Soal SIMAK UI 2012 [Soal Lengkap Disini]$ \underset{x \to \infty}{lim} \left ( 5^{x}+5^{3x} \right )^{\frac{1}{x}}= \cdots $.
$\begin{align}
(A)\ & 0 \\
(B)\ & 1 \\
(C)\ & 10 \\
(D)\ & 25 \\
(E)\ & 125
\end{align}$Alternatif Pembahasan:

Hint

Soal ini ada keunikan alasannya rumus-rumus di atas tidak ada membahas yang menyerupai ini, jadi untuk soal ini kita coba dengan manipulasi aljabar;
$\begin{align}
& \underset{x \to \infty}{lim} \left( 5^{x}+5^{3x} \right)^{\frac{1}{x}} \\
& =\underset{x \to \infty}{lim} \left( 5^{3x} \left( \dfrac{1}{5^{2x}}+1 \right) \right)^{\frac{1}{x}} \\
& =\underset{x \to \infty}{lim} \left( 5^{3x} \right)^{\frac{1}{x}} \left( \dfrac{1}{5^{2x}}+1 \right)^{\frac{1}{x}}\\
& =\underset{x \to \infty}{lim} 5^{3} \left( \dfrac{1}{5^{2x}}+1 \right)^{\frac{1}{x}}\\
& = 125 \left( \dfrac{1}{5^{\infty}}+1 \right)^{\frac{1}{\infty}}\\
& = 125 \left( 0 +1 \right)^{0} \\
& = 125
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(E)\ 125$

6. Soal SIMAK UI 2010 [Soal Lengkap Disini]$ \underset{x \to \infty}{lim} \dfrac{2^{x+1}-3^{x-2}+4^{x+1}}{2^{x-1}-3^{x+1}+4^{x-1}}= \cdots $.
$\begin{align}
(A)\ & \dfrac{1}{16} \\
(B)\ & \dfrac{1}{8} \\
(C)\ & \dfrac{1}{4} \\
(D)\ & 16 \\
(E)\ & 32
\end{align}$Alternatif Pembahasan:

Hint

Soal ini ada keunikan alasannya rumus-rumus di atas dan yang dibahas di buku-buku tidak ada membahas yang menyerupai ini, jadi untuk soal ini kita coba dengan manipulasi aljabar;
$\begin{align}
& \underset{x \to \infty}{lim} \dfrac{2^{x+1}-3^{x-2}+4^{x+1}}{2^{x-1}-3^{x+1}+4^{x-1}} \\
& =\underset{x \to \infty}{lim} \dfrac{2^{x} \cdot 2^{1} - 3^{x} \cdot 3^{-2}+ 4^{x} \cdot 4^{1}}{2^{x} \cdot 2^{-1}-3^{x} \cdot 3^{1}+4^{x} \cdot 4^{-1}} \\
& =\underset{x \to \infty}{lim} \dfrac{2^{x} \cdot 2 - 3^{x} \cdot \frac{1}{9} + 4^{x} \cdot 4}{2^{x} \cdot \frac{1}{2} -3^{x} \cdot 3+4^{x} \cdot \frac{1}{4}} \cdot \dfrac{\frac{1}{4^{x}}}{\frac{1}{4^{x}}} \\
& =\underset{x \to \infty}{lim} \dfrac{\dfrac{2^{x}}{4^{x}} \cdot 2 - \dfrac{3^{x}}{4^{x}} \cdot \dfrac{1}{9} + \dfrac{4^{x}}{4^{x}} \cdot 4}{\dfrac{2^{x}}{4^{x}} \cdot \dfrac{1}{2} - \dfrac{3^{x}}{4^{x}} \cdot 3+ \dfrac{4^{x}}{4^{x}} \cdot \dfrac{1}{4}} \\
& =\underset{x \to \infty}{lim} \dfrac{\dfrac{1}{2^{x}} \cdot 2 - \left (\dfrac{3}{4} \right )^{x} \cdot \dfrac{1}{9} + 4}{\dfrac{1}{2^{x}} \cdot \dfrac{1}{2} - \left (\dfrac{3}{4} \right )^{x} \cdot 3+ \dfrac{1}{4}} \\
& =\dfrac{\dfrac{1}{\infty} \cdot 2 - \left (\dfrac{3}{4} \right )^{\infty} \cdot \dfrac{1}{9} + 4}{\dfrac{1}{\infty} \cdot \dfrac{1}{2} - \left (\dfrac{3}{4} \right )^{\infty} \cdot 3+ \dfrac{1}{4}} \\
& =\dfrac{0 \cdot 2 - 0 \cdot \dfrac{1}{9} + 4}{0 - 0 + \dfrac{1}{4}} \\
& =\dfrac{4}{\dfrac{1}{4}}=16
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(D)\ 16$

7. Soal SIMAK UI 2010 [Soal Lengkap Disini]$\underset{x \to \infty}{lim} \left ( \sqrt{x^2+3x}- x \right )=\cdots$
$\begin{align}
(A)\ & -1 \\
(B)\ & 0 \\
(C)\ & \dfrac{3}{2} \\
(D)\ & 3 \\
(E)\ & \infty
\end{align}$Alternatif Pembahasan:

Hint

$\begin{align}
& \underset{x \to \infty}{lim} \left ( \sqrt{x^2+3x}- x \right ) \\
& =\underset{x \to \infty}{lim} \left ( \sqrt{x^2+3x}- \sqrt{x^2} \right ) \\
& =\frac{b-q}{2\sqrt{a}} \\
& =\frac{3-0}{2\sqrt{1}} \\
& =\frac{3}{2}
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(C)\ \frac{3}{2}$

8. Soal SIMAK UI 2010 [Soal Lengkap Disini]$\underset{x \to \infty}{lim} \left ( \left ( 2\sqrt{x}+1 \right ) - \sqrt{4x - 3\sqrt{x}+2} \right )=\cdots$
$\begin{align}
(A)\ & - \infty \\
(B)\ & -\dfrac{3}{4} \\
(C)\ & 1 \\
(D)\ & \dfrac{7}{4} \\
(E)\ & \infty
\end{align}$Alternatif Pembahasan:

Hint

$\begin{align}
& \underset{x \to \infty}{lim} \left ( \left ( 2\sqrt{x}+1 \right ) - \sqrt{4x - 3\sqrt{x}+2} \right ) \\
& = \underset{x \to \infty}{lim} \left ( \sqrt{\left ( 2\sqrt{x}+1 \right )^{2}} - \sqrt{4\sqrt{x} - 3\sqrt{x}+2} \right ) \\
& = \underset{x \to \infty}{lim} \left ( \sqrt{ \left (2\sqrt{x} \right )^{2}+2 \left (2\sqrt{x} \right )+1} - \sqrt{\left (2\sqrt{x} \right )^{2} - \frac{3}{2} \left (2\sqrt{x} \right )+2} \right ) \\
& \text{misal:}\ 2\sqrt{x}=m,\ \text{karena}\ x \to \infty\ \text{maka}\ m \to \infty \\
& = \underset{m \to \infty}{lim} \left ( \sqrt{ m^{2}+2 m+1} - \sqrt{m^{2} - \frac{3}{2} m+2} \right ) \\
& =\frac{b-q}{2\sqrt{a}} \\
& =\frac{2+\frac{3}{2}}{2\sqrt{1}} \\
& =\frac{\frac{7}{2}}{2}=\frac{7}{4}
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(D)\ \frac{7}{4}$

9. Soal SBMPTN 2017 [Soal Lengkap Disini]$\underset{x \to \infty}{lim} \dfrac{x^{4}\ sin \left (\dfrac{1}{x}\right )+x^{2}}{1+x^{3}}=\cdots$
$\begin{align}
(A)\ & \text{Tidak ada limitnya} \\
(B)\ & 0 \\
(C)\ & 1 \\
(D)\ & - \infty \\
(E)\ & \infty
\end{align}$Alternatif Pembahasan:

Hint

Untuk menuntaskan soal limit ini kita gunakan sedikit manipulasi aljabar dan bekerjasama sedikit dengan limit fungsi trigonometri.
Misalkan $\dfrac{1}{x}=m$ maka $\dfrac{1}{m}=x$. Karena $x \to \infty$ maka $m \to 0$.

Soal $\underset{x \to \infty}{lim} \dfrac{x^{4}\ sin\left (\dfrac{1}{x} \right )+x^{2}}{1+x^{3}}$ dapat kita tuliskan menjadi
$\begin{align}
& \underset{m \to 0}{lim} \dfrac{\left (\dfrac{1}{m} \right )^{4}\ sin\ m+\left (\dfrac{1}{m} \right )^{2}}{1+\left (\dfrac{1}{m} \right )^{3}} \\
& = \underset{m \to 0}{lim} \dfrac{\dfrac{1}{m^{4}}\ sin\ m+\dfrac{1}{m^{2}}}{1+\dfrac{1}{m^{3}}} \\
& = \underset{m \to 0}{lim} \dfrac{\dfrac{sin\ m}{m^{4}}+\dfrac{1}{m^{2}}}{1+\dfrac{1}{m^{3}}} \cdot \dfrac{m^{3}}{m^{3}} \\
& = \underset{m \to 0}{lim} \dfrac{\dfrac{sin\ m}{m}+m}{m^{3}+1} \\
& = \dfrac{1+0}{0+1}\\
& = 1
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(C)\ 1$

10. Soal STIS 2017 $\underset{x \to \infty}{lim} \left (2+ \dfrac{2}{2+\dfrac{2}{2+\dfrac{2}{x}}} \right )=\cdots$
$\begin{align}
(A)\ & \dfrac{3}{8} \\
(B)\ & 2 \\
(C)\ & \dfrac{8}{3} \\
(D)\ & 3 \\
(E)\ & \infty
\end{align}$Alternatif Pembahasan:

Hint

$\begin{align}
& \underset{x \to \infty}{lim} \left (2+ \dfrac{2}{2+\dfrac{2}{2+\dfrac{2}{x}}} \right ) \\
& =\underset{x \to \infty}{lim} \left (2+ \dfrac{2}{2+\dfrac{2}{2+\dfrac{2}{\infty}}} \right ) \\
& = \left (2+ \dfrac{2}{2+\dfrac{2}{2+0}} \right ) \\
& = \left (2+ \dfrac{2}{2+1} \right ) \\
& = \left (2+ \dfrac{2}{3} \right ) \\
& = \left (2 \dfrac{2}{3} \right ) \\
& = \dfrac{8}{3}
\end{align}$

$\therefore$ Pilihan yang sesuai ialah $(C)\ \dfrac{8}{3}$

"Jika engkau tidak mampu menahan lelahnya belajar, Maka engkau harus menanggung pahitnya kebodohan" ___pythagoras

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