1.次の極限を求めなさい。
(1)\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}\)
相加平均・相乗平均より
\(\displaystyle x^2y^2\leqq\left(\frac{x^2+y^2}{2}\right)^2\)
すなわち
\(\displaystyle 0\leqq\frac{x^2y^2}{x^2+y^2}\leqq\frac{(x^2+y^2)^2}{4(x^2+y^2)}\)
\(\displaystyle 0\leqq\frac{x^2y^2}{x^2+y^2}\leqq\frac{x^2+y^2}{4}\)
ここで
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{4}=0\)
よって、はさみうちの定理より
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0\)
(2)\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2y^2(x-1)^2(y-1)^2}{x^2(x-1)^2+y^2(y-1)^2}\)
相加平均・相乗平均より
\(\displaystyle x^2y^2(x-1)^2(y-1)^2\leqq\left(\frac{x^2(x-1)^2+y^2(y-1)^2}{2}\right)^2\)
すなわち
\(\displaystyle 0\leqq\frac{x^2y^2(x-1)^2(y-1)^2}{x^2(x-1)^2+y^2(y-1)^2}\)
\(\displaystyle \ \ \ \leqq\frac{x^2(x-1)^2+y^2(y-1)^2}{4}\)
ここで
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2(x-1)^2+y^2(y-1)^2}{4}=0\)
よって、はさみうちの定理より
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2y^2(x-1)^2(y-1)^2}{x^2(x-1)^2+y^2(y-1)^2}=0\)
(3)\(\displaystyle \lim_{(x,y)\to(1,1)}\frac{x^2y^2(x-1)^2(y-1)^2}{x^2(x-1)^2+y^2(y-1)^2}\)
相加平均・相乗平均より
\(\displaystyle x^2y^2(x-1)^2(y-1)^2\leqq\left(\frac{x^2(x-1)^2+y^2(y-1)^2}{2}\right)^2\)
すなわち
\(\displaystyle 0\leqq\frac{x^2y^2(x-1)^2(y-1)^2}{x^2(x-1)^2+y^2(y-1)^2}\)
\(\displaystyle \ \ \ \leqq\frac{x^2(x-1)^2+y^2(y-1)^2}{4}\)
ここで
\(\displaystyle \lim_{(x,y)\to(1,1)}\frac{x^2(x-1)^2+y^2(y-1)^2}{4}=0\)
よって、はさみうちの定理より
\(\displaystyle \lim_{(x,y)\to(1,1)}\frac{x^2y^2(x-1)^2(y-1)^2}{x^2(x-1)^2+y^2(y-1)^2}=0\)
(4)\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x-y}{x+y}\)
\(x=0\)として\((x,y)\to(0,0)\)のとき、
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x-y}{x+y}=\frac{-y}{y}=-1\)
\(y=0\)として\((x,y)\to(0,0)\)のとき、
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x-y}{x+y}=\frac{x}{x}=1\)
よって、
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x-y}{x+y}\)は存在しない。
(5)\(\displaystyle \lim_{(x,y)\to(0,0)}xy\log(x^2+y^2)\)
相加平均・相乗平均より
\(\displaystyle |xy\log(x^2+y^2)|\leqq\left|\frac{x^2+y^2}{2}\log(x^2+y^2)\right|\)
すなわち
\(\displaystyle 0\leqq|xy\log(x^2+y^2)|\leqq\left|\frac{x^2+y^2}{2}\log(x^2+y^2)\right|\)
ここで
\(\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{2}|\log(x^2+y^2)|=0\)
よって、はさみうちの定理より
\(\displaystyle \lim_{(x,y)\to(0,0)}xy\log(x^2+y^2)=0\)
