01　関数の極限

• 関数 \(\displaystyle f(x) = \frac{\sin{x}}{x}\) は \(x \ne 0\) で定義された関数です。
• したがって，\(f(0)\) を考えることはできません。
• しかし，\(x\) の値を限りなく \(0\) に近づけることはできます。
• そして，そのとき \(1\) に限りなく近づく（収束する）ということです。

1. \(\displaystyle \lim_{x \to 1}\frac{x - 1}{\sqrt{x} - 1}\)
2. \(\displaystyle \lim_{x \to \infty} \frac{4x^2 + 5x - 3}{3x^2 + x + 1}\)
3. \(\displaystyle \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right) \)

1. \(\mbox{Limit((x - 1) / (sqrt(x) - 1), x, 1)}\)
2. \(\mbox{Limit((4*x^2 + 5*x -3) / (3*x^2 + x + 1), x, Infinity)}\)
3. \(\mbox{Limit(sqrt(x^2 + 1) - x, x, Infinity)}\)

\(x\)

\(y\)

\(\displaystyle y = \frac{1}{x}\)

1. \(\displaystyle \lim_{x \to 2} \frac{x^2 - 5x + 6}{x - 2}\)　解答 隠す

\(\displaystyle \lim_{x \to 2} \frac{x^2 - 5x + 6}{x - 2} \begin{array}[t]{l} \displaystyle = \lim_{x \to 2} \frac{\cancel{(x - 2)}(x -3 )}{\cancel{(x - 2)}} \\ \displaystyle = \lim_{x \to 2} (x - 3) \\ = -1 \end{array}\)

2. \(\displaystyle \lim_{x \to -1} \frac{2x^2 - 3x - 5}{3x^2 + 4x + 1}\)　解答 隠す

\(\displaystyle \lim_{x \to -1} \frac{2x^2 - 3x - 5}{3x^2 + 4x + 1}\begin{array}[t]{l} \displaystyle = \lim_{x \to -1} \frac{\cancel{(x + 1)}(2x - 5)}{\cancel{(x + 1)}(3x + 1)} \\ \displaystyle = \lim_{x \to -1} \frac{2x - 5}{3x + 1} \\ \displaystyle = \frac{7}{2} \end{array}\)

3. \(\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x + 2}\)　解答 隠す

\(\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x + 2}\begin{array}[t]{l} \displaystyle = \lim_{x \to -2} \frac{\cancel{(x + 2)}(x^2 - 2x + 4)}{\cancel{x + 2}} \\ \displaystyle = \lim_{x \to -2} \left(x^2 - 2x + 4\right) \\ = 12 \end{array}\)

4. \(\displaystyle \lim_{x \to 1} \frac{x^3 + 2x^2 - 2x - 1}{x^2 - 1}\)　解答 隠す

\(\displaystyle \lim_{x \to 1} \frac{x^3 + 2x^2 - 2x - 1}{x^2 - 1}\begin{array}[t]{l} \displaystyle = \lim_{x \to 1} \frac{\cancel{(x - 1)}(x^2 + 3x + 1)}{\cancel{(x - 1)}(x + 1)} \\ \displaystyle = \lim_{x \to 1} \frac{x^2 + 3x + 1}{x + 1} \\ \displaystyle = \frac{5}{2} \end{array}\)

5. \(\displaystyle \lim_{x \to 3} \frac{x^3 - 6x^2 + 11x - 6}{x^3 - 7x - 6}\)　解答 隠す

\(\displaystyle \lim_{x \to 3} \frac{x^3 - 6x^2 + 11x - 6}{x^3 - 7x - 6}\begin{array}[t]{l} \displaystyle = \lim_{x \to 3} \frac{\cancel{(x - 3)}(x - 2)(x - 1)}{\cancel{(x - 3)}(x + 2)(x + 1)} \\ \displaystyle = \lim_{x \to 3} \frac{(x - 2)(x - 1)}{(x + 1)(x + 2)} \\ \displaystyle = \frac{1}{10} \end{array}\)

6. \(\displaystyle \lim_{x \to 0} \frac{\sin^2{x}}{1 - \cos{x}}\)　解答 隠す

\(\displaystyle \lim_{x \to 0} \frac{\sin^2{x}}{1 - \cos{x}}\begin{array}[t]{l} \displaystyle = \lim_{x \to 0} \frac{1 - \cos^2 x}{1 - \cos x} \\ \displaystyle = \lim_{x \to 0} \frac{\cancel{(1 - \cos x)}(1 + \cos x)}{\cancel{1 - \cos x}} \\ \displaystyle = \lim_{x \to 0} (1 + \cos x) \\ = 2 \end{array}\)

7. \(\displaystyle \lim_{x \to \infty} \frac{3x^2 + 2x + 1}{x^2 + 2x + 3}\)　解答 隠す

\(\displaystyle \lim_{x \to \infty} \frac{3x^2 + 2x + 1}{x^2 + 2x + 3}\begin{array}[t]{l} \displaystyle = \lim_{x \to \infty} \frac{3 + \frac{2}{x} + \frac{1}{x^2}}{1 + \frac{2}{x} + \frac{3}{x^2}} \\ = 3 \end{array}\)

8. \(\displaystyle \lim_{x \to \infty} \frac{x^3 + 1}{x^2 + 3x + 2}\)　解答 隠す

\(\displaystyle \lim_{x \to \infty} \frac{x^3 + 1}{x^2 + 3x + 2}\begin{array}[t]{l} \displaystyle = \lim_{x \to \infty} \frac{x + \frac{1}{x^2}}{1 + \frac{3}{x} + \frac{2}{x^2}} \\ = \infty \end{array}\)

9. \(\displaystyle \lim_{x \to \infty} \frac{10x^2 + 5x - 3}{x^3 + 3x^2 + 3x + 1}\)　解答 隠す

\(\displaystyle \lim_{x \to \infty} \frac{10x^2 + 5x - 3}{x^3 + 3x^2 + 3x + 1}\begin{array}[t]{l} \displaystyle = \lim_{x \to \infty} \frac{\frac{10}{x} + \frac{5}{x^2} - \frac{3}{x^3}}{1 + \frac{2}{x} + \frac{3}{x^2} + \frac{1}{x^3}} \\ = 0 \end{array}\)

10. \(\displaystyle \lim_{x \to -\infty} \frac{x^3 - x + 6}{x^2 + 4x + 2}\)　解答 隠す

\(\displaystyle \lim_{x \to -\infty} \frac{x^3 - x + 6}{x^2 + 4x + 2}\begin{array}[t]{l} \displaystyle = \lim_{x \to -\infty} \frac{x - \frac{1}{x} + \frac{6}{x^2}}{1 + \frac{4}{x} + \frac{2}{x^2}} \\ = -\infty \end{array}\)

11. \(\displaystyle \lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x - 2}\)　解答 隠す

\(\displaystyle \lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x - 2}\begin{array}[t]{l} \displaystyle = \lim_{x \to 2} \frac{(\sqrt{x + 7} - 3)(\sqrt{x + 7} + 3)}{(x - 2)(\sqrt{x + 7} + 3)} \\ \displaystyle = \lim_{x \to 2} \frac{(x + 7) - 9}{(x - 2)(\sqrt{x + 7} + 3)} \\ \displaystyle = \lim_{x \to 2} \frac{\cancel{(x - 2)}}{\cancel{(x - 2)}(\sqrt{x + 7} + 3)} \\ \displaystyle = \lim_{x \to 2} \frac{1}{\sqrt{x + 7} + 3} \\ \displaystyle = \frac{1}{6} \end{array}\)

12. \(\displaystyle \lim_{x \to 1} \frac{x - 1}{\sqrt{x} - \sqrt{2 - x}}\)　解答 隠す

\(\displaystyle \lim_{x \to 1} \frac{x - 1}{\sqrt{x} - \sqrt{2 - x}}\begin{array}[t]{l} \displaystyle = \lim_{x \to 1} \frac{(x - 1)(\sqrt{x} + \sqrt{2 - x})}{(\sqrt{x} - \sqrt{2 - x})(\sqrt{x} + \sqrt{2 - x})} \\ \displaystyle = \lim_{x \to 1} \frac{(x - 1)(\sqrt{x} + \sqrt{2 - x})}{x - (2 - x)} \\ \displaystyle = \lim_{x \to 1} \frac{\cancel{(x - 1)}(\sqrt{x} + \sqrt{2 - x})}{2\cancel{(x - 1)}} \\ \displaystyle = \lim_{x \to 1} \frac{\sqrt{x} + \sqrt{2 -x}}{2} \\ = 1 \end{array}\)

13. \(\displaystyle \lim_{x \to -1} \frac{\sqrt{x + 2} + x}{\sqrt{x + 5} - 2}\)　解答 隠す

\(\begin{array}[t]{l} \displaystyle \displaystyle \lim_{x \to -1} \frac{\sqrt{x + 2} + x}{\sqrt{x + 5} - 2} \\ \displaystyle = \lim_{x \to -1} \frac{(\sqrt{x + 2} + x)(\sqrt{x + 2} - x)(\sqrt{x + 5} + 2)}{(\sqrt{x + 5} - 2)(\sqrt{x + 5} + 2)(\sqrt{x + 2} - x)} \\ \displaystyle = \lim_{x \to -1} \frac{\{(x + 2) - x^2\}(\sqrt{x + 5} + 2)}{\{(x + 5) - 4\}(\sqrt{x + 2} - x)} \\ \displaystyle = \lim_{x \to -1} \frac{-\cancel{(x + 1)}(x - 2)(\sqrt{x + 5} + 2)}{\cancel{(x + 1)}(\sqrt{x + 2} - x)} \\ \displaystyle = \lim_{x \to -1} \frac{-(x - 2)(\sqrt{x + 5} + 2)}{\sqrt{x + 2} - x} \\ = 6 \end{array}\)

14. \(\displaystyle \lim_{x \to -1 - 0} \frac{1}{x + 1}\)　解答 隠す

\(\displaystyle \lim_{x \to -1 - 0} \frac{1}{x + 1}\begin{array}[t]{l} \displaystyle = -\infty \end{array}\)

15. \(\displaystyle \lim_{x \to \frac{\pi}{2} + 0} \tan x\)　解答 隠す

\(\displaystyle \lim_{x \to \frac{\pi}{2} + 0} \tan x\begin{array}[t]{l} = -\infty \end{array}\)

16. \(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1}{\tan x}\)　解答 隠す

\(\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{1}{\tan x} \begin{array}[t]{l} = 0 \end{array}\)

17. \(\displaystyle \lim_{x \to 0} x\sin\frac{1}{x}\)　解答 隠す

\(\begin{array}[t]{l} \displaystyle -1 \leqq \sin\frac{1}{x} \leqq 1 \\ \displaystyle \mbox{∴}\quad -|\,x\,| \leqq x\sin\frac{1}{x} \leqq |\,x\,| \\ \displaystyle \lim_{x \to 0} |\,x\,| = 0 \quad {\small より}\quad \lim_{x \to 0} x\sin\frac{1}{x} = 0 \end{array}\)

18. \(\displaystyle \lim_{x \to \infty} 2^{-x}\sin x\)　解答 隠す

\(\begin{array}[t]{l} \displaystyle -1 \leqq \sin x \leqq 1 \\ \displaystyle \mbox{∴}\quad -2^{-x} \leqq 2^{-x}\sin x \leqq 2^{-x} \\ \displaystyle \lim_{x \to \infty} 2^{-x} = 0 \quad {\small より}\quad \lim_{x \to \infty} 2^{-x}\sin x = 0 \end{array}\)