Bank Soal Limit Aljabar Dan Trigonometri

Beberapa koleksi soal yang sering ditemui dalam ujian ujian tes masuk universitas ini aku bahas. Mungkin dapat menjadi soal rujukan ulangan dan aku rasa soal dan pembahasan limit trigonometri dan aljabar ini ini cukup gampang untuk dikerjakan. Selamat memahami.

Soal 1. $ \lim_{x \rightarrow 3} \frac {x^2-9}{ \sqrt {10+2x}-x-1}=...$
$\lim_{x \rightarrow 3} \frac {x^2-9}{ \sqrt {10+2x}-x-1} \\ \lim_{x \rightarrow 3} \frac {x^2-9}{ \sqrt {10+2x}-(x+1)} \\ \text {Kalikan akar sekawan} \\ \lim_{x \rightarrow 3} \frac {x^2-9}{ \sqrt {10+2x}-(x+1)}. \frac {\sqrt {10+2x}+(x+1)}{\sqrt {10+2x}+(x+1)} \\ \lim_{x \rightarrow 3} \frac {x^2-9}{ 9-x^2}  = -1$

Soal 2.  $\lim_{x \rightarrow \infty } \sqrt {25x^2-9x+6} -5x+1=...$
$\lim_{x \rightarrow \infty } \sqrt {25x^2-9x+6} -5x+1= \lim_{x \rightarrow \infty } \sqrt {25x^2-9x+6} - (5x-1) \\ \lim_{x \rightarrow \infty } \sqrt {25x^2-9x+6} - \sqrt {(5x-1)^2} \\ \lim_{x \rightarrow \infty } \sqrt {25x^2-9x+6} - \sqrt {25x^2-10x+1} $
Penyelesaian
$ \frac {b-q}{2 \sqrt a} \\ \frac {-9-(-10)}{2 \sqrt {25}} = \frac {1}{10}$

Soal 3. $\lim_{x \rightarrow \infty } \sqrt {4x^2-2x+1} +\sqrt {x^2+x+6} -3x +6 =...$
$\lim_{x \rightarrow \infty } \sqrt {4x^2-2x+1} +\sqrt {x^2+x+6} -3x +6 \\ \lim_{x \rightarrow \infty } \sqrt {4x^2-2x+1} +\sqrt {x^2+x+6} -2x -x +6 \\ \lim_{x \rightarrow \infty } \sqrt {4x^2-2x+1} +\sqrt {x^2+x+6}   +6 \\ \lim_{x \rightarrow \infty } \sqrt {4x^2-2x+1} -\sqrt {4x^2} +\sqrt {x^2+x+6} -\sqrt {x^2} +6 \\ \frac {-2-0}{2 \sqrt 4 }+ \frac {1-0}{\sqrt 1} +6 =6,5$

Soal 4. $\lim_{x \rightarrow 1 } \frac {(x^2-1) \sin 2 (x-1)}{-2 \sin ^2 (x-1)}=...$
$\lim_{x \rightarrow 1 } \frac {(x^2-1) \sin 2 (x-1)}{-2 \sin ^2 (x-1)} \\ \lim_{x \rightarrow 1 } \frac {(x-1)(x+1) \sin 2 (x-1)}{-2 \sin  (x-1) \sin  (x-1)} \\ \lim_{x \rightarrow 1 } \frac {(x+1)2}{-2} =-2$

Soal 5. $\lim_{x \rightarrow 2 } \frac {x^3 -4x}{x-2} =...$
$\lim_{x \rightarrow 2 } \frac {x^3 -4x}{x-2} \\ \lim_{x \rightarrow 2 } \frac {x (x-2)(x+2)}{x-2} \\ \lim_{x \rightarrow 2 } x(x+2) =8$

Soal 6. $\lim_{x \rightarrow 0 } \frac {2x \sin 3x}{1- \cos 6x}=... $

$ \lim_{x \rightarrow 0 } \frac {2x \sin 3x}{1- \cos 6x} \\ \lim_{x \rightarrow 0 } \frac {2x \sin 3x}{1- (1-2 \sin ^2 3x)} \\ \lim_{x \rightarrow 0 } \frac {2x \sin 3x}{2 \sin 3x \sin 3x} =\frac {1}{3}$

Soal 7. $\lim_{x \rightarrow 0 } \frac {1- \cos 2x}{x \tan \frac {1}{2}x} =...$

$\lim_{x \rightarrow 0 } \frac {1- \cos 2x}{x \tan \frac {1}{2}x} \\ \lim_{x \rightarrow 0 } \frac {1- (1 -2 \sin^2x)}{x \tan \frac {1}{2}x} \\ \lim_{x \rightarrow 0 } \frac {2 \sin x \sin x}{x \tan \frac {1}{2}x} =4$

Soal 8. $ \lim_{x \rightarrow 0 } \frac {\cos 2x}{\cos x - \sin x}=...$

$\lim_{x \rightarrow 0 } \frac {\cos 2x}{\cos x - \sin x} \\ \lim_{x \rightarrow 0 } \frac {\cos ^2x - \sin ^2x}{\cos x - \sin x} \\  \lim_{x \rightarrow 0 } \frac {(\cos x - \sin x)(\cos x + \sin x)}{\cos x - \sin x} \\  \lim_{x \rightarrow 0 } (\cos x + \sin x) = 1  $

Soal 9. $\lim_{x \rightarrow 0 } \frac {5x \tan x}{1- \cos 6x } =...$

$\lim_{x \rightarrow 0 } \frac {5x \tan x}{1- \cos 6x } \\ \lim_{x \rightarrow 0 } \frac {5x \tan x}{1- (1- 2\sin ^2 3x)}  \\ \lim_{x \rightarrow 0 } \frac {5x \tan x}{2\sin 3x  \sin 3x  } = \frac {5}{18}$

Soal 10. $\lim_{x \rightarrow 0 } \frac {\cos 4x -1}{\cos x -1 }=...$

$\lim_{x \rightarrow 0 } \frac {\cos 4x -1}{\cos x -1 } \\ \lim_{x \rightarrow 0 } \frac {\cos 4x - \cos 0}{\cos x - \cos 0} \\ \lim_{x \rightarrow 0 } \frac {-2 \sin 2x \sin 2x} {-2 \sin \frac {1}{2}x \sin \frac {1}{2}x } =8$

Soal 11. $ \lim_{x \rightarrow 0 } \frac {\cos 5x -\cos x}{\cos 3x - \cos x }=...$

$ \lim_{x \rightarrow 0 } \frac {\cos 5x -\cos x}{\cos 3x - \cos x } \\ \lim_{x \rightarrow 0 } \frac {-2 \sin 3x \sin 2x}{-2 \sin 2x \sin x } =3$

Soal 12 $\lim_{x \rightarrow 0 } \frac {\sin 3x +\sin x}{\tan 3x }=... $

$\lim_{x \rightarrow 0 } \frac {\sin 3x +\sin x}{\tan 3x } \\ \lim_{x \rightarrow 0 } \frac {\sin 2x . \cos x}{\tan 3x } \\ \lim_{x \rightarrow 0 } \frac {3 . \cos x}{3 } = \frac {2}{3}$

Soal 13 . $\lim_{x \rightarrow 0 } \frac {\sin 2x -1}{\sin 10x -1 } =... $

Turunkan pembilang dan penyebut.

$\lim_{x \rightarrow 0 } \frac {\sin 2x -1}{\sin 10x -1 } \\ \lim_{x \rightarrow 0 } \frac { 2 \cos 2x}{10 \cos 10x } = \frac {1}{5}$

Catatan lain:

Anda harus ingat rumus penjumlahan trigonometri

$\lim_{x \rightarrow 0 } \frac {\tan ax }{\tan bx } \\ \lim_{x \rightarrow 0 }  \frac {\tan ax }{\sin bx } \\  \lim_{x \rightarrow 0 } \frac {\sin ax }{\tan bx } \\  \lim_{x \rightarrow 0 }  \frac {\sin ax }{bx } \\ \lim_{x \rightarrow 0 }   \frac {\sin ax }{\sin bx } \\ \lim_{x \rightarrow 0 }   \frac { ax }{\sin bx } \\ \text {Nilainya: } \frac {a}{b}$

Sumber http://www.marthamatika.com/

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