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14\frac{\text{1}}{\text{4}}41
12\frac{\text{1}}{\text{2}}21
13\frac{\text{1}}{\text{3}}31
1
−1-1−1
−2-2−2
−3-3−3
12\frac{1}{2}21
0
13\frac{1}{3}31
111
-1
Tentukan nilai lim dari limx→0 sin 3x . tan 5x1−cos 5x\lim_{x\rightarrow0}\ \frac{\sin\ 3x\ .\ \tan\ 5x}{1-\cos\ 5x}x→0lim 1−cos 5xsin 3x . tan 5x adalah . . .
25\frac{2}{5}52
35\frac{3}{5}53
45\frac{4}{5}54
65\frac{6}{5}56
limx→ 14π cos 2xcos x − sin x=\lim_{x\rightarrow\ \frac{1}{4}\pi}\ \frac{\cos\ 2x}{\cos\ x\ -\ \sin\ x}=x→ 41πlim cos x − sin xcos 2x=
2\sqrt{2}2
12 2\frac{1}{2}\ \sqrt{2}21 2
14 2\frac{1}{4}\ \sqrt{2}41 2
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