limit as (x,y) approaches (0,0) of (e^{xy}-1)/(xy)

解

(x, y) \to (0, 0) lim (\frac{e ^{x y} - 1}{x y})

1

解答ステップ

(x, y) \to (0, 0) lim (\frac{e ^{x y} - 1}{x y})

極座標に変換する

x = r cos (θ)

= r \to 0 lim (\frac{e ^{r c o s (θ) r s i n (θ)} - 1}{r cos ( θ ) r sin ( θ )})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{cos ( θ ) sin ( θ )} \cdot r \to 0 lim (\frac{e ^{r c o s (θ) r s i n (θ)} - 1}{rr})

簡素化

\frac{e ^{r c o s (θ) r s i n (θ)} - 1}{rr} : \frac{e ^{r^{2} c o s (θ) s i n (θ)} - 1}{r ^{2}}

= \frac{1}{cos ( θ ) sin ( θ )} \cdot r \to 0 lim (\frac{e ^{r^{2} c o s (θ) s i n (θ)} - 1}{r ^{2}})

ロピタルの定理を適用する

= \frac{1}{cos ( θ ) sin ( θ )} \cdot r \to 0 lim (\frac{e ^{r^{2} c o s (θ) s i n (θ)} r sin ( 2 θ )}{2 r})

簡素化

\frac{e ^{r^{2} c o s (θ) s i n (θ)} r sin ( 2 θ )}{2 r} : \frac{1}{2} e^{r^{2} c o s (θ) s i n (θ)} sin (2 θ)

= \frac{1}{cos ( θ ) sin ( θ )} \cdot r \to 0 lim (\frac{1}{2} e^{r^{2} c o s (θ) s i n (θ)} sin (2 θ))

値を挿入する

r = 0

= \frac{1}{cos ( θ ) sin ( θ )} \cdot \frac{1}{2} e^{0^{2} c o s (θ) s i n (θ)} sin (2 θ)

簡素化

\frac{1}{cos ( θ ) sin ( θ )} \cdot \frac{1}{2} e^{0^{2} c o s (θ) s i n (θ)} sin (2 θ) : 1

= 1