Question

Вычислить пределы функций, не пользуясь дифференциального средствами исчисления. ​

Answer 1

Ответ:

Пошаговое объяснение:

1)

$\displaystyle \lim_{x \to 1} \frac{3x^2-4x+1}{\sqrt{8+x}-3} = \lim_{x \to 1} \frac{(3x^2-4x+1)*(\sqrt{8+x}+3)}{(\sqrt{8+x}-3)(\sqrt{8+x}+3)} =\\ = \lim_{x \to 1} \frac{(3x^2-4x+1)*(\sqrt{8+x}+3)}{8+x-9} = \lim_{x \to 1} \frac{3(x-1)(x-\frac{1}{3})*(\sqrt{8+x}+3)}{x-1} = \lim_{x \to 1} (3x-1)(\sqrt{8+x}+3) = (3 - 1)(\sqrt{9} + 3) = 2 * 6 = 12$

2)

$\displaystyle \lim_{x \to \infty} \frac{\sqrt[4]{x^8 + 1} + x}{\sqrt{x^4 + 2}} = \left[ : x^2 \right] =\lim_{x \to \infty} \frac{\sqrt[4]{1 + \frac{1}{x^8}} + \frac{1}{x}}{\sqrt{1 + \frac{2}{x^4}}} = \left[ x \to \infty, \frac{1}{x} \to 0 \right] = \\= \lim_{x \to \infty} \frac{\sqrt[4]{1 }}{\sqrt{1}} =1$

3)

$\displaystyle \lim_{x \to 0} \frac{\sqrt{sinx}}{\sqrt{x+1} - 1} = \left[ x \to 0, \ \sin x \sim x \right] = \lim_{x \to 0} \frac{\sqrt{x}}{\sqrt{x+1} - 1} =\\ = \lim_{x \to 0} \frac{\sqrt{x}(\sqrt{x +1} + 1)}{x} = \lim_{x \to 0} \frac{\sqrt{x +1} + 1}{\sqrt{x}} = \lim_{x \to 0} \left(\sqrt{1 +\frac{1}{x}} + \frac{1}{\sqrt{x}}}\right) = \infty$

4)

$\displaystyle \lim_{x \to \infty} \left(\frac{2x^2+5}{2x^2+3}\right)^{-x^2-2} = \lim_{x \to \infty} \left(1 + \frac{2}{2x^2+3}\right)^{-x^2-2} =\\ = \lim_{x \to \infty} \left(1 + \frac{1}{x^2+1.5}\right)^{-x^2-2} = \lim_{x \to \infty} \left(1 + \frac{1}{x^2+1.5}\right)^{(x^2+1.5)\frac{-(x^2+2)}{x^2+1.5}} =\\ = \left[ \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e \right] = \lim_{x \to \infty} e^{\cfrac{-(x^2+2)}{x^2+1.5}} = \lim_{x \to \infty} e^{\cfrac{-(1+\frac{2}{x^2})}{1+\frac{1.5}{x^2}}} =$$= \displaystyle \left[ x \to \infty, \ \frac{1}{x} \to 0 \right] = e^{-1} = \frac{1}{e}$

5)

$\displaystyle \lim_{x \to \infty} x(\ln (x+5) - ln x) = \lim_{x \to \infty} x\left(\ln \frac{x+5}{x}\right) = \lim_{x \to \infty} x\left(\ln \left(1+\frac{5}{x}\right)\right) =\\ = \lim_{x \to \infty} x\left(\ln \left(1+\frac{1}{0.2x}\right)^{0.2x * \frac{5}{x}}\right) = \lim_{x \to \infty} x\ln e^{\frac{5}{x}} = \lim_{x \to \infty} x * \frac{5}{x}\ln e = \lim_{x \to \infty} 5\ln e =\\ =5$