Question

Помогите с решением)) ​

Answer 1

Ответ:

1. -9

2. -2

3. 0

4. 3

5. $e^{\frac{7}{6}}$

Пошаговое объяснение:

1. $\lim_{x \to \ 2} \frac{2x^{2}+x-10}{x^{2}-5x+6}=\\=\lim_{x \to \ 2} \frac{(2x+5)(x-2)}{(x-3)(x-2)}=\\=\lim_{x \to \ 2} \frac{2x+5}{x-3}=\frac{2*2+5}{2-3}=-9$

2. $\lim_{x \to \ -1} \frac{1-\sqrt{x+2}}{2-\sqrt{3-x}}$

x+2=t², x --> -1  ⇒  t² --> 1 ⇒ x=t²-2, t --> 1

$\lim_{t \to \ 1} \frac{1-t}{2-\sqrt{3-(t^{2}-2)}}=\lim_{t \to \ 1} \frac{1-t}{2-\sqrt{5-t^{2}}}=\\=\lim_{t \to \ 1} \frac{(1-t)*(2+\sqrt{5-t^{2}})}{(2-\sqrt{5-t^{2}})*(2+\sqrt{5-t^{2}})}=\\=\lim_{t \to \ 1} \frac{(1-t)*(2+\sqrt{5-t^{2}})}{4-(5-t^{2})}=\\=\lim_{t \to \ 1} \frac{(1-t)*(2+\sqrt{5-t^{2}})}{t^{2}-1}=\\=\lim_{t \to \ 1} \frac{(1-t)*(2+\sqrt{5-t^{2}})}{(t-1)*(t+1)}=\\=\lim_{t \to \ 1} \frac{-(2+\sqrt{5-t^{2}})}{t+1}=\frac{-(2+\sqrt{5-1^{2}})}{1+1}=\\=\frac{-(2+2)}{2}=-2$

3. $\lim_{x \to \infty} \frac{x^{5}-2x^{2}+3x^{3}}{x^{6}-7x}= \\=\lim_{x \to \infty} \frac{\frac{x^{5}-2x^{2}+3x^{3}}{x^{6}}}{\frac{x^{6}-7x}{x^{6}}}=\\= \lim_{x \to \infty} \frac{\frac{1}{x}-\frac{2}{x^{4}}+\frac{3}{x^{3}}}{1-\frac{7}{x^{5}}}=\\=\frac{0-0+0}{1-0}=0$

4. Используем $\lim_{x \to \ 0} \frac{sinax}{ax}=1$

$\lim_{x \to \ 0} \frac{tg5x-sin2x}{x}=\lim_{x \to \ 0} \frac{tg5x}{x}-\lim_{x \to \ 0} \frac{sin2x}{x}=\\=\lim_{x \to \ 0} \frac{\frac{sin5x}{cos5x}}{x}-\lim_{x \to \ 0} \frac{2*sin2x}{2*x}=\\=\lim_{x \to \ 0} \frac{5*sin5x}{5*x*cos5x}-2*1=\\=5*1-2=3$

5. Используем $\lim_{a \to \infty} (1+\frac{1}{a})^{a}=e$

$\lim_{x \to \ 0} (1+\frac{2x}{3})^{\frac{7}{4x}}=$

$y=\frac{1}{x}, {x \to \ 0}$ ⇒ $\ y \to \infty}$

$= \lim_{y \to \infty}(1+\frac{2}{3y})^{\frac{7y}{4}}=\\= \lim_{y \to \infty}((1+\frac{2}{3y})^{\frac{3y}{2}})^{\frac{2*7}{3*4}}=\\= \lim_{y \to \infty}((1+\frac{2}{3y})^{\frac{3y}{2}})^{\frac{7}{6}}=\\=e^{\frac{7}{6}}$