関数の極限(limit of function)

演習問題

1.: 次の極限値を求めよう．
(a): $\displaystyle{\lim_{x \rightarrow 2}(x^2 + 4x)}$
(b): $\displaystyle{\lim_{x \rightarrow 2}\frac{x^2 - 3x + 2}{x^2 -4}}$
(c): $\displaystyle{\lim_{x \rightarrow 0}\frac{2x^4 - 6x^3 + x^2 + 2}{x - 1}}$
(d): $\displaystyle{\lim_{x \rightarrow 1}\frac{2x^4 - 6x^3 + x^2 + 3}{x - 1}}$
(e): $\displaystyle{\lim_{x \rightarrow 0}\frac{\sqrt{1+x} - \sqrt{1 - x}}{x}}$
(f): $\displaystyle{\lim_{x \rightarrow 0}\frac{(1+x)^{1/3} - (1 - x)^{1/3}}{x}}$
(g): $\displaystyle{\lim_{x \rightarrow 2} f(x)}$ , ただし $\displaystyle{f(x) = \left\{\begin{array}{cl} x^2, & x \neq 2\\ 3, & x = 2 \end{array}\right.}$
2.: $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin{x}}{x} = 1}$ を用いて次の極限値を求めよう．
(a): $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin{2x}}{\sin{3x}}}$
(b): $\displaystyle{\lim_{x \rightarrow 0}\frac{\cos{x} - 1}{x}}$
(c): $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin^{-1}{x}}{x}}$
(d): $\displaystyle{\lim_{x \rightarrow 0}x \sin\frac{1}{x}}$
3.: $\lim_{x \rightarrow a}\vert f(x)\vert = 0$ ならば， $\lim_{x \rightarrow a}f(x) = 0$ を示そう．