与えられた極限の計算問題を解きます。 (11) $\lim_{x\to 0} \frac{x^3+2x^2}{2x^4-3x^2}$ (12) $\lim_{x\to \infty} \frac{2x+3}{x^2-3}$ (13) $\lim_{x\to \infty} \frac{3x^2+1}{2x^2+x+1}$ (14) $\lim_{x\to 0} \frac{\sqrt{1+x}-1}{x}$ (15) $\lim_{x\to 0} \frac{\sin 3x}{x}$ (16) $\lim_{x\to 0} \frac{\tan 2x}{x}$ (17) $\lim_{x\to 0} \frac{1-\cos 2x}{x^2}$ (18) $\lim_{x\to \infty} (1+\frac{2}{x})^x$ (19) $\lim_{x\to 0} (1+3x)^{\frac{1}{x}}$ (20) $\lim_{x\to 0} \frac{\log_e(1-x)}{x}$

はい、承知いたしました。画像にある極限の計算問題を解きます。

1. 問題の内容

与えられた極限の計算問題を解きます。

(11) $\lim_{x\to 0} \frac{x^3+2x^2}{2x^4-3x^2}$

(12) $\lim_{x\to \infty} \frac{2x+3}{x^2-3}$

(13) $\lim_{x\to \infty} \frac{3x^2+1}{2x^2+x+1}$

(14) $\lim_{x\to 0} \frac{\sqrt{1+x}-1}{x}$

(15) $\lim_{x\to 0} \frac{\sin 3x}{x}$

(16) $\lim_{x\to 0} \frac{\tan 2x}{x}$

(17) $\lim_{x\to 0} \frac{1-\cos 2x}{x^2}$

(18) $\lim_{x\to \infty} (1+\frac{2}{x})^x$

(19) $\lim_{x\to 0} (1+3x)^{\frac{1}{x}}$

(20) $\lim_{x\to 0} \frac{\log_e(1-x)}{x}$

2. 解き方の手順

(11)

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

(12)

$\lim_{x\to \infty} \frac{2x+3}{x^2-3} = \lim_{x\to \infty} \frac{\frac{2}{x}+\frac{3}{x^2}}{1-\frac{3}{x^2}} = \frac{0+0}{1-0} = 0$

(13)

$\lim_{x\to \infty} \frac{3x^2+1}{2x^2+x+1} = \lim_{x\to \infty} \frac{3+\frac{1}{x^2}}{2+\frac{1}{x}+\frac{1}{x^2}} = \frac{3+0}{2+0+0} = \frac{3}{2}$

(14)

$\lim_{x\to 0} \frac{\sqrt{1+x}-1}{x} = \lim_{x\to 0} \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \lim_{x\to 0} \frac{1+x-1}{x(\sqrt{1+x}+1)} = \lim_{x\to 0} \frac{x}{x(\sqrt{1+x}+1)} = \lim_{x\to 0} \frac{1}{\sqrt{1+x}+1} = \frac{1}{\sqrt{1+0}+1} = \frac{1}{1+1} = \frac{1}{2}$

(15)

$\lim_{x\to 0} \frac{\sin 3x}{x} = \lim_{x\to 0} \frac{\sin 3x}{3x} \cdot 3 = 1 \cdot 3 = 3$

(16)

$\lim_{x\to 0} \frac{\tan 2x}{x} = \lim_{x\to 0} \frac{\tan 2x}{2x} \cdot 2 = 1 \cdot 2 = 2$

(17)

$\lim_{x\to 0} \frac{1-\cos 2x}{x^2} = \lim_{x\to 0} \frac{2\sin^2 x}{x^2} = 2 \lim_{x\to 0} (\frac{\sin x}{x})^2 = 2(1)^2 = 2$

(18)

$\lim_{x\to \infty} (1+\frac{2}{x})^x = e^2$

(19)

$\lim_{x\to 0} (1+3x)^{\frac{1}{x}} = \lim_{x\to 0} e^{\frac{1}{x}\ln(1+3x)} = e^{\lim_{x\to 0} \frac{\ln(1+3x)}{x}} = e^{\lim_{x\to 0} \frac{3/(1+3x)}{1}} = e^3$

(20)

$\lim_{x\to 0} \frac{\log_e(1-x)}{x} = \lim_{x\to 0} \frac{\ln(1-x)}{x} = \lim_{x\to 0} \frac{-1/(1-x)}{1} = \lim_{x\to 0} \frac{-1}{1-x} = \frac{-1}{1-0} = -1$

3. 最終的な答え

(11) $-\frac{2}{3}$

(12) $0$

(13) $\frac{3}{2}$

(14) $\frac{1}{2}$

(15) $3$

(16) $2$

(17) $2$

(18) $e^2$

(19) $e^3$

(20) $-1$