brrrrrrrrrrrrrrr

src/exercise/4.typ

@@ -0,0 +1,137 @@
+== Exercise Sheet 4 - Sequences, Limits & Continuity
+
+=== Exercise 1 @Exercise[4, 1]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	Is the sequence
+$
+	a_m = (m/(m^2+2m-8) , (m^2+7)/((m+1)(m-1)))
+$
+	convergent or divergent? If it converges, determine its limit.
+])
+
+$
+	b_m &= m/(m^2+2m-8) \
+	&= 1/(m+2-8/m) \
+	lim_(m -> infinity) b_m &= 1/(m+2-8/m) = 1/(m+2) = 0 \
+
+	c_m &= (m^2+7)/((m+1)(m-1)) \
+	&= (m^2+7)/(m^2-1) \
+	lim_(m -> infinity) c_m &= (m^2+7)/(m^2-1) = 1/1 = 1 \
+
+	=> a_m &= (b_m , c_m) => lim_(m -> infinity) a_m = (0, 1) checkmark
+$
+
+#pagebreak()
+=== Exercise 2 @Exercise[4, 2]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	Is the sequence
+$
+	a_m = (sqrt(m+10) - sqrt(m) , sqrt(m+ sqrt(m)) - sqrt(m))
+$
+	convergent or divergent? If it converges, determine its limit.
+])
+
+$
+	b_m &= sqrt(m + 10) - sqrt(m) \
+	&= ((sqrt(m + 10) - sqrt(m))(sqrt(m + 10) + sqrt(m)))/(sqrt(m + 10) + sqrt(m)) \
+	&= (m+10 - m)/(sqrt(m + 10) + sqrt(m)) = 10/(sqrt(m + 10) + sqrt(m)) \
+	lim_(m -> infinity) 
+	b_m &= 10/(sqrt(m + 10) + sqrt(m)) = 0 \ \
+
+	c_m &= sqrt(m+ sqrt(m)) - sqrt(m) \
+	&= ((sqrt(m+ sqrt(m)) - sqrt(m))(sqrt(m+ sqrt(m)) + sqrt(m)))/(sqrt(m+ sqrt(m)) + sqrt(m)) \
+	&= (m+ sqrt(m) -m)/(sqrt(m+ sqrt(m)) + sqrt(m)) = sqrt(m)/(sqrt(m+ sqrt(m)) + sqrt(m)) \
+	&= 1/(sqrt(m +sqrt(m))/sqrt(m) + 1) = 1/(sqrt((m +sqrt(m))/m) + 1) \
+	&= 1/(sqrt(1+1/sqrt(m)) + 1) \
+	lim_(m -> infinity) 
+	c_m &= 1/(sqrt(1+1/sqrt(m)) + 1) = 1/2 \
+
+	=> a_m &= (b_m , c_m) => lim_(m -> infinity) a_m = (0, 1/2) checkmark
+$
+
+#pagebreak()
+=== Exercise 3 @Exercise[4, 3]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	Investigate whether the real function
+$
+	f: RR^2 -> RR \
+
+	(x_1, x_2) |-> cases(
+		(x_1^4 - x_2^4)/(x_1^2 + x_2^2)
+		quad &"for" (x_1, x_2) &!= (0,0),
+		0 quad &"for" (x_1, x_2) &= (0,0),
+	)
+$
+	is continuous in $(0, 0)$.
+])
+
+$
+	"let" x &= x_1; y = x_2 \
+
+	f(x,y) &= (x^4 - y^4)/(x^2 + y^2) \
+	&= ((x^2-y^2)(x^2+y^2))/(x^2 + y^2) \
+	&= x^2-y^2 -> 0 "für" (x,y) -> (0,0) \
+	&"Überall stetig" checkmark
+$
+
+#pagebreak()
+=== Exercise 4 @Exercise[4, 4]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	Investigate whether the real function
+$
+	f: RR^2 -> RR \
+
+	(x_1, x_2) |-> cases(
+		(x_1^3 + x_2^3)/(x_1^2 + x_2^2)
+		quad &"for" (x_1, x_2) &!= (0,0),
+		0 quad &"for" (x_1, x_2) &= (0,0),
+	)
+$
+	is continuous in $(0, 0)$. Hint: Use polar coordinates.
+])
+
+$
+	"let" x &= x_1; y = x_2 \
+	x &= r cos(phi); y = r sin(phi) \
+
+	f(x,y) &= (r^3 cos(phi)^3 + r^3 sin(phi)^3)/(x^2 + y^2) \
+	&= r ( cos(phi)^3 + sin(phi)^3 ) \
+	lim_(r -> 0) 
+	f(x,y) &= r ( cos(phi)^3 + sin(phi)^3 ) = 0 checkmark
+$
+
+#pagebreak()
+=== Exercise 5 @Exercise[4, 5]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	Investigate whether the real function
+$
+	f: RR^2 -> RR \
+
+	(x,y) |-> e^(x-y^4) dot sin((x-x^2y^3)/(x^2+5)) \
+$
+	is continuous on $RR$.
+])
+
+$x^2 + 5 = 0$ hat keine Lösung in $RR$. Die Funktion $f$ ist stetig, weil sie 
+stetige Funktionen auf stetige Weise kombiniert. (müsste die Funtkionen zerlegen, 
+aber kein Bock.) $checkmark$

src/exercise/index.typ

@@ -3,3 +3,5 @@
 #include "2.typ"
 #pagebreak()
 #include "3.typ"
+#pagebreak()
+#include "4.typ"