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Soal 19

\(\displaystyle \lim_{x \rightarrow \sim} \: \dfrac {3x^6 - 5x^2}{(1 - 3x^3)^2}\)

 


Soal 31

\(\displaystyle \lim_{x \rightarrow 2} \: \dfrac {x^2 - 4}{x - 2}\)

 


Soal 33

\(\displaystyle \lim_{x \rightarrow 4} \: \dfrac {x^2 - 4x}{x - 4}\)

 


\(\displaystyle \lim_{x \rightarrow \sim} \: \frac {(2x^2 - 7)^3 \:.\: (6x^2 + 2x)}{(3 + 4x^3) \:.\: (3x^5 + 7x)} = \dotso\)

(A)   \(2\)

(B)   \(4\)

(C)   \(6\)

(D)   \(8\)

(E)   \(10\)

 

\(\displaystyle \lim_{x \rightarrow 2} \: \frac {3x - 6}{x - 3} = \dotso\)

(A)   \(0\)

(B)   \(1\)

(C)   \(2\)

(D)   \(3\)

(E)   \(4\)

Jawab: D

 

\begin{equation*} \begin{split} & \lim_{x \rightarrow 2} \: \frac {3x - 6}{x - 2} \\\\ & \lim_{x \rightarrow 2} \: \frac {3 \cancel{(x - 2)}}{\cancel{x - 2}} \\\\ & 3 \end{split} \end{equation*}


 

Mathematics (Prev Lesson)

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