SOAL SBMPTN LIMIT FUNGSI ALJABAR DAN PEMBAHASAN - Kikiholmes

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1. 

   1. 3
   2. $\frac{3}{2}$
   3. $\sqrt{3}$
   4. 1
   5. 0
Jawaban B





2. 

   1. -6
   2. -4
   3. -1
   4. 4
   5. 6
Jawaban E





3. 

   1. $-\frac{1}{3}$
   2. $-\frac{1}{5}$
   3. 0
   4. $\frac{1}{5}$
   5. $\frac{1}{3}$
Jawaban D





4. \[\begin{align*} \text{Jika }\lim_{x\to a} \left (f \left ( x \right )+\frac{1}{g\left ( x \right )} \right )=4\\ \text{ dan } \lim_{x\to x} \left (f \left ( x \right )-\frac{1}{g\left ( x \right )} \right )=-3,\\ \text{Maka }\lim_{x\to a}\left (\left (f \left ( x \right ) \right )^2+ \left ( \frac{1}{g\left (x \right )} \right )^2 \right )= \cdots \end{align*}\]

   1. $\frac{24}{3}$
   2. $\frac{23}{5}$
   3. $\frac{25}{3}$
   4. $\frac{25}{2}$
   5. $\frac{27}{2}$
Jawaban D

\[\begin{align*} \bullet \lim_{x\to a} \left (f \left ( x \right )+\frac{1}{g\left ( x \right )} \right )=4\\ \\ \text{kedua ruas dikuadratkan}\\ \\ \lim_{x\to a} f^2 \left(x \right )+\lim_{x\to a} \left (\frac{1}{g\left (x \right )} \right )^2 +2 \lim_{x\to a} f\left(x \right) \cdot \frac{1}{g \left(x \right)} &=4^2 \ \ \cdots \left(i \right) \\ \\ \bullet \lim_{x\to a} \left (f\left (x \right )-\frac{1}{g\left (x \right )} \right ) &=-3\\ \\ \text{Kedua ruas juga dikuadratkan }\\ \\ \lim_{x\to a}f^2 \left (x \right ) + \lim_{x\to a}\left (\frac{1}{g\left (x \right )} \right)^2 \lim_{x\to a}f\left (x \right ) \cdot \frac{1}{g\left (x \right )} &=\left (-3 \right )^2\ \ \cdots \left (ii \right ) \end{align*}\] \[\begin{array}{l} \text{Persamaan (i) dan (ii) dijumlahkan, maka diperoleh,}\\ 2 \left \{ \lim_{x\to a} f^2 \left ( x \right )+\lim_{x\to a}\left ( \frac{1}{g\left ( x \right )} \right )^2 \right \} =25\\ \left \{ \lim_{x\to a} f^2 \left ( x \right )+\lim_{x\to a}\left ( \frac{1}{g\left ( x \right )} \right )^2 \right \} = \frac{25}{2}\\ \end{array}{}\]



5. Nilai dari $\displaystyle \lim_{x \to \infty} \frac{8x^3-6x^2+5x-3}{4x^3-2x^2+3x+2}=\cdots$

   1. $-4$
   2. $-2$
   3. $-\frac{1}{4}$
   4. 2
   5. 4
Jawaban D

\[\begin{align*} \text{Karena pangkat tertinggi pembilang}\\ \text{ sama dengan pangkat tertinggi penyebut} \text{dan sesuai dengan bentuk }\\ \lim_{x\to \infty}\frac{a_1x^m+a_2x^{m-1}+\cdots}{b_1x^n+b_2x^{n-1}+\cdots}&=\frac{a_1}{b_1},\\ \\ \text{jika m=n, maka}\\ \lim_{x\to\infty} \frac{8x^3-6x^2+5x-3}{4x^3-2x^2+3x+2} &=\frac{8}{4}\\ &= 2 \end{align*}\]



6. Nilai dari $\displaystyle \lim_{x\to 4} \frac{x^3-64}{x^2-5x+4}=\cdots$

   1. 64
   2. 16
   3. 4
   4. -4
   5. -16
Jawaban B

\[\begin{align*} \lim_{x\to4} \frac{x^3-64}{x^2-5x+4}&= \lim_{x\to4}\frac{\left (x-4 \right )\left (x^2+4x+16 \right )}{\left (x-4 \right ) \left (x-1 \right )}\\ &= \lim_{x\to4}\frac{x^2+4x+16}{x-1}\\ &= \frac{4^2+4\cdot4+16}{4-1}\\ &= 16 \end{align*}\]



7. Jika $\displaystyle \lim_{x\to 0} \frac{g\left( x \right )}{x}=\frac{1}{2}$,
   maka nilai $\displaystyle\lim_{x\to 0} \frac{g\left( x\right )}{\sqrt{1-x}-1}=\cdots$

   1. -4
   2. -2
   3. -1
   4. 2
   5. 4
Jawaban C

$$\displaystyle \begin{aligned}\lim_{x\to 0} \frac{g\left( x\right )}{\sqrt{1-x}-1}\\ &= \lim_{x\to 0} \frac{g\left ( x \right )}{\sqrt{1-x}-1}\cdot \lim_{x\to 0} \frac{\sqrt{1-x}+1}{\sqrt{1-x}+1}\\ &=\lim_{x\to 0}\frac{g\left ( x \right )\left \{ \sqrt{1-x}+1 \right \}}{\left( 1-x \right )-1}\\ &=\lim_{x\to 0}\frac{g\left(x \right )\left \{ \sqrt{1-x}+1 \right \}}{-x}\\ &=-\lim_{x\to 0}\frac{g\left ( x \right )}{x}\cdot\lim_{x\to 0}\left ( \sqrt{1-x}+1 \right )\\ &=-\frac{1}{2}\cdot \left(\sqrt{1-0}+1 \right )\\ &=-\frac{1}{2}\cdot 2\\ &=-1\end{aligned}$$



8. $\displaystyle\text{Nilai } \lim_{x\to 0}\frac{\sqrt{4x}}{\sqrt{\sin2x}}=\cdots$

   1. $\sqrt{2}$
   2. 1
   3. 0
   4. -1
   5. -2
Jawaban A

$$\displaystyle \begin{aligned} \lim_{x\to 0}\frac{\sqrt{4x}}{\sqrt{\sin2x}}\\ &=\left ( \lim_{x\to 0} \frac{2\cdot 2x}{\sin2x} \right )^{\frac{1}{2}}\\ &=\left ( \lim_{x\to 0} \frac{2x}{\sin2x}\cdot \lim_{x\to 0}2 \right )^\frac{1}{2}\\ &=\left ( 1\cdot \lim_{x\to 0}2 \right )^\frac{1}{2}\\ &=\left( 2 \right )^\frac{1}{2}\\ &=\sqrt{2} \end{aligned}$$



9. Nilai $\displaystyle \lim_{x\to 0}\frac{\sqrt[3]{1+cx}-1}{x}=\cdots$

   1. $\frac{1}{3}c$
   2. $\frac{2}{3}c$
   3. c
   4. 2c
   5. 0
Jawaban A

$$\displaystyle \begin{aligned}\lim_{x\to 0}\frac{\sqrt[3]{1+cx}-1}{x}\\ &=\lim_{x\to 0}\frac{\frac{1}{3}c\cdot \frac{1}{\sqrt[3]{\left (1+cx \right )^2}}}{1}\ \ \ \rightarrow \boxed{\text{dengan cara L'Hospital}}\\&=\frac{1}{3}c\cdot \frac{1}{\sqrt[3]{\left ( 1+c\left (0 \right ) \right )^2}}\\ &=\frac{1}{3}c \end{aligned}$$



10. $$\displaystyle \lim_{x\to1}\left [ \left ( \frac{4}{x^2-x}-\frac{4-3x+x^2}{1-x^3} \right )^{-1}+\frac{4\left ( x^4-1 \right )}{x^2-x^{-1}} \right ]=\cdots$$

    1. 0
    2. $\frac{4}{3}$
    3. $\frac{8}{3}$
    4. $\frac{16}{3}$
    5. $\infty$
Jawaban D

$$\displaystyle \begin{aligned} &=\lim_{x\to1}\left [ \left ( \frac{4}{x^2-x}-\frac{4-3x+x^2}{1-x^3} \right )^{-1}+\frac{4\left ( x^4-1 \right )}{x^2-x^{-1}} \right ]\\ &=\lim_{x\to1}\left [ \left ( \frac{4}{x\left ( x-1 \right )}+\frac{4-3x+x^2}{\left ( x-1 \right )\left ( \left ( x-1 \right )^2 +3x \right )} \right )^{-1}+\frac{4\left (x^2+1 \right )\left ( x^2-1 \right )}{x^2-\frac{1}{x}} \right ]\\ &=\lim_{x\to1}\left [ \left ( \frac{4}{x\left ( x-1 \right )}+\frac{4-3x+x^2}{\left ( x-1 \right )\left ( x^2+x+1 \right )} \right )^{-1}+\frac{4x\left ( x^2+1 \right )\left ( x+1 \right )\left ( x-1 \right )}{x^3-1} \right ]\\ &=\lim_{x\to1}\left [ \left ( \frac{4\left ( x^2+x+1 \right )+\left ( 4-3x+x^2 \right )\left ( x \right )}{x \left ( x-1 \right )\left ( x^2+x+1 \right )} \right )^{-1}+\frac{4x\left ( x^2+1 \right )\left ( x+1 \right )\left ( x-1 \right )}{\left ( x-1 \right )\left ( x^2+x+1 \right )} \right ]\\ &=\lim_{x\to1}\left [ \left ( \frac{x^3+x^2+8x+4}{x\left ( x-1 \right )\left ( x^2+x+1 \right )} \right )^{-1}+\frac{4x\left ( x^2+1 \right )\left ( x+1 \right )}{\left ( x^2+x+1 \right )} \right ]\\ &=\lim_{x\to1}\left [ \frac{x\left ( x-1 \right )\left ( x^2+x+1 \right )}{x^3+x^3+8x+4}+\frac{4x\left ( x^2+1 \right )\left ( x+1 \right )}{x^2+x+1} \right ]\\ &=\lim_{x\to1}\left [ \frac{1\left ( 1-1 \right )\left ( 1^2+1+1 \right )}{1^3+1^3+8\cdot 1+4}+\frac{4\cdot 1\left ( 1^2+1 \right )\left ( 1+1 \right )}{1^2+x+1} \right ]\\ &=0+\frac{16}{3}\\ &=\frac{16}{3} \end{aligned}$$

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