Question

Решите

1) lim 3x1²+x-4/5-3x-2x²

x→1

2) lim 3х²-4х+1/6х²+5х-3

x→∞

3) lim sin(1-2x) /1-4x²

x→½

4) lim (2n+3)/(2n+1)^n+1

n→∞

Answer 1

$1) \ \lim_{x \to 1} \dfrac{3x^{2} + x - 4}{5-3x-2x^{2}}=\bigg\{\dfrac{0}{0} \bigg\} = \lim_{x \to 1} -\dfrac{(3x+4)(x-1)}{(2x+5)(x-1)}= \\= \lim_{x \to 1} -\dfrac{3x+4}{2x+5} = -\dfrac{3 \cdot 1 + 4}{2 \cdot 1 + 5} = -1$

$2) \ \lim_{x \to \infty} \dfrac{3x^{2} - 4x + 1}{6x^{2} + 5x - 3} = \bigg\{\dfrac{\infty}{\infty} \bigg\} =\begin{vmatrix}3x^{2} - 4x + 1 \sim 3x^{2} \\ 6x^{2} + 5x - 3 \sim 6x^{2}\end{vmatrix} = \lim_{x \to \infty} \dfrac{3x^{2}}{6x^{2}} = \dfrac{1}{2}$

$3) \ \lim_{x \to \frac{1}{2} } \dfrac{\sin(1-2x)}{1-x^{4}}= \dfrac{\bigg(1 - 2\cdot \dfrac{1}{2} \bigg)}{1 - \dfrac{1}{16} } =\dfrac{0}{\dfrac{15}{16} } = 0$

$4) \ \lim_{n \to \infty} \bigg(\dfrac{2n+3}{2n+1} \bigg)^{n+1} =\{1^{\infty} \} = \lim_{n \to \infty} \bigg(\dfrac{2n+1 + 2}{2n+1} \bigg)^{\frac{2n+2}{2}} =\\=\lim_{n \to \infty} \bigg(1+\dfrac{2}{2n+1} \bigg)^{\frac{2n+1+1}{2}}=\lim_{n \to \infty} \bigg(1+\dfrac{2}{2n+1} \bigg)^{\frac{2n+1}{2}+\frac{1}{2} }=\\=\lim_{n \to \infty} \bigg(1+\dfrac{2}{2n+1} \bigg)^{\frac{2n+1}{2}} \cdot \lim_{n \to \infty} \bigg(1+\dfrac{2}{2n+1} \bigg)^{\frac{1}{2}}=e \cdot 1^{1/2} = e$