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

Lösung

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

1

Schritte zur Lösung

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

Schreibe in Polarkoordinaten um

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})

Vereinfache

\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}})

Wende das L'Hopital Theorem an

= \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})

Vereinfache

\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))

Setze den Wert

r = 0

ein

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

Vereinfache

\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

\frac{\partial}{\partial x} (x^{4} + 3 y^{3})

x \to 1 lim (x^{3} - 3 x + 2)

\int \frac{1}{x + x ^{2}} d x

y^{^{''}} + \frac{1}{2} y = 0

d er i v a t i v e f (x) = (5 x + 9) x^{2}