Question

1.lim 7x сверху 2 ctg3x/ arcsin 2x

2. lim 7(1-cos2x)ctg3x/arcsin 2x

Answer 1

Ответ:

$1)\ \ \lim\limits_{x \to 0}\dfrac{7x^2\, ctg3x}{arcsin2x}=\lim\limits_{x \to 0}\ \dfrac{7x^2}{tg3x\cdot arcsin2x}=\lim\limits_{x \to 0}\ \dfrac{7x^2}{3x\cdot 2x}=\dfrac{7x^2}{6x^2}=\dfrac{7}{6}$

$2)\ \ \lim\limits_{x \to 0}\dfrac{7(1-cos2x)\cdot ctg3x}{arcsin2x}=\lim\limits_{x \to 0}\ \dfrac{7\cdot 2sinx}{tg3x\cdot arcsin2x}=\lim\limits_{x \to 0}\dfrac{7\cdot 2x}{3x\cdot 2x}=\\\\\\=\lim\limits_{x \to 0}\ \dfrac{14x}{6x^2}=\lim\limits_{x \to 0}\dfrac{7}{3x}=\Big[\dfrac{7}{0}\Big]=\infty$