heureka

(without using l'hopital's rule) how to do you prove that:

lim x->0 (tan3x)/x =3 ? 

$$\small{\text{
$
\boxed{ \lim\limits_{x\to0}
\left(
\dfrac{ \tan{(3x)} } { x } \right)
\qquad \tan{ (3x) } &= (3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb
} % boxed
$
}}$\\\\\\$
\small{\text{
$
\lim\limits_{x\to0}\left(
\dfrac{ \tan{(3x)} } { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
(3x)+\dfrac13 (3x)^3+\dfrac{2}{15}(3x)^5+\dfrac{17}{315}(3x)^7+\dotsb
} { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
x \left( 3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right)
} { x } \right)
$
}}$\\\\\\$
\small{\text{
$
=
\lim\limits_{x\to0}\left(
3+\dfrac13 3^3x^2+\dfrac{2}{15}3^5x^4+\dfrac{17}{315}3^7x^6+\dotsb \right)= \textcolor[rgb]{1,0,0}{3}
$
}}$$

(without using l'hopital's rule) how to do you prove that:

lim x->0 (sinx)/x=1 ? 

$$\small{\text{
$
\boxed{ \lim\limits_{x\to0}
\left(
\dfrac{ \sin{(x)} } { x } \right)
\qquad \sin (x) = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!} = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}\mp\dotsb
} % boxed
$
}}$\\\\\\$
\small{\text{
$
\lim\limits_{x\to0}\left(
\dfrac{ \sin{(x)} } { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
\dfrac{x}{1!}-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}\mp\dotsb
} { x } \right)
=
\lim\limits_{x\to0}\left(
\dfrac{
x \left( \dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right)
} { x } \right)
$
}}$\\\\\\$
\small{\text{
$
=
\lim\limits_{x\to0}\left(
\dfrac{1}{1!}-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}\mp\dotsb \right)= \dfrac{1}{1!} = \textcolor[rgb]{1,0,0}{1}
$
}}$$

Now its just the last one that I cant figure out if its right or wrong:

z = a+bi --> r and θ (degrees)

z = conjugate(4+2i) = 4-2i

$$\small{\text{$\boxed{r=\sqrt{a^2+b^2}\qquad\phi= \left\{ \begin{array}{ll}\arctan{( \frac{b}{a} )}, & a > 0 \text{ and } b \text{ whatever!} \\ \\\arctan{( \frac{b}{a} )} + \pi , & a < 0 \text{ and } b \ge 0 \\\\\arctan{( \frac{b}{a} )} - \pi , & a < 0 \text{ and } b < 0 \\\\\frac{ \pi }{2}, & a = 0 \text{ and } b > 0 \\\\-\frac{ \pi }{2}, & a = 0 \text{ and } b < 0 \\\\\end{array}\right.}$}}$$

$$\\\small{\text{\ $z = 4-2i$}}\\
\small{\text{ $a=4, \ b = -2$}}\\
\small{\text{ $r=\sqrt{(4)^2+(-2)^2} = \sqrt{20}=4.47$}} \\
\small{\text{ $a > 0 :\quad$ $\phi= \arctan{( \frac{b}{a} )} =\arctan{( \frac{-2}{4} )} = -26.5650511771\ensurement{^{\circ}} $}}$$

3. also final velocity v: 

$$v=v_0+at_{\small{\text{stone hits the ground}}} \quad v_0 = 7\quad t = 6.51625939587s \\
v = 7 + 32 * 6.51625939587 = 215.5203 \ \frac{ft}{s}\\$$

What is the common ratio sequence 6,-3,1.5,-0.75

$$\small{\text{the common ratio is
$\quad \frac{-3}{6}= \frac{1.5}{-3}= \frac{-0.75}{1.5}= -\frac{1}{2} $
\quad \small{\text{or}}\quad $-0.5$
}}$$

The path of a fired projectile is modeled by the function f(x)=x(2x+8) where x is the height (in feet) and f(x) in the distance(in feet). If the distance is 24 feet, what is the height of the projectile ?

$$f(x)=x(2x+8)\quad f(x) = 24 \\
24 = x(2x+8) \\
24 = 2x^2+8x \quad | \quad :2\\
12 = x^2 + 4x \quad | \quad -12\\
0 = x^2 + 4x -12 \\
x^2 + 4x -12 = 0 \quad \small{\text{factoring}}\\
\underbrace{(x-2)}_{=0}\underbrace{(x+6)}_{=0} =0 \\
x-2=0 \quad \small{\text{we have} }
\quad x=2 \quad \small{\text{okay}} \\
x+6=0 \quad \small{\text{we have}} \quad x=-6 \quad \small{\text{no solution!}}\\$$

the height of the projectile is 2 feet.

Rewrite to polar form re ^ iθ, where θ is an angle in radians

$$\small{\text{$\boxed{z=a+bi \quad \to \quad re^{i\phi} }$}}\\\\
\small{\text{
$
\boxed{r=\sqrt{a^2+b^2}
\qquad
\phi= \left\{
\begin{array}{ll}
\arctan{( \frac{b}{a} )}, & a > 0 \\ \\
\arctan{( \frac{b}{a} )} + \pi , & a < 0 \text{ and } b \ge 0 \\\\
\arctan{( \frac{b}{a} )} - \pi , & a < 0 \text{ and } b < 0 \\\\
 \frac{ \pi }{2}, & a = 0 \text{ and } b > 0 \\\\
-\frac{ \pi }{2}, & a = 0 \text{ and } b < 0 \\\\
\end{array}
\right.
}
$
}}$$

I.

   $$\\\small{\text{\ $z = -3$}}\\
\small{\text{ $a=-3, \ b = 0 $}}\\
\small{\text{ $r=\sqrt{(-3)^2+0^2} = 3$}} \\
\small{\text{ $a < 0, \ b=0:\quad$ $\phi=\arctan{( \frac{b}{a} )} + \pi
=\arctan{( \frac{0}{-3} )}+\pi = 0+\pi = \pi
$}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = -3 = 3e^{i\pi}}$}}$$

II.

$$\\\small{\text{\ $z = 1-i$}}\\
\small{\text{ $a=1, \ b = -1$}}\\
\small{\text{ $r=\sqrt{(1)^2+(-1)^2} = \sqrt{2}$}} \\
\small{\text{ $a > 0, \ b < 0:\quad$ $\phi= \arctan{( \frac{b}{a} )} =\arctan{( \frac{-1}{1} )} = -\frac{\pi}{4}
$}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = 1-i = \sqrt{2}e^{-\frac{\pi}{4}
i}}$}}$$

III.

$$\\\small{\text{\ $z = -5i$}}\\
\small{\text{ $a=0, \ b = -5$}}\\
\small{\text{ $r=\sqrt{(0)^2+(-5)^2} = 5$}} \\
\small{\text{ $a = 0, \ b < 0:\quad$ $\phi= -\frac{ \pi }{2} $}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = -5i = 5e^{-\frac{ \pi }{2} i}}$}}$$

Using the Gauss method to add 1+2+3+...+98+99+100= 50(1+100)=5050,

Find the sum of  1x+2+3x+4+5x+...+98+99x+100.

Can you please show how you got the answer as well, thank you.

$$\small{\text{
$
1x+2+3x+4+5x+ \dots +98+99x+100 =
$
}}\\
\small{\text{
$
\underbrace{
[\ 1x+3x+5x+7x+ \dots + 97x+99x \ ]
}_{\text{Part}\ 1}
+
\underbrace{
[ 2+4+6+8+\dots +98+100\ ]
}_{\text{Part}\ 2}
$
}}$$

Part 1:

$$\small{\text{
$[\ 1x+3x+5x+7x+ \dots + 97x+99x \ ]
=
2x[\ \underbrace{ 1+2+3+4+\dots + 49+ 50}_{=(1+50)*\frac{50}{2}}\ ] -50x $
}}\\
\small{\text{
$=(1+50) *50x-50x = 2550x-50x = 2500x
$
}}$$

Part 2:

$$\small{\text{
$
[ 2+4+6+8+\dots +98+100\ ] = 2*[\ \underbrace{ 1+2+3+4+\dots + 49+ 50}_{=(1+50)*\frac{50}{2}}\ ]
$
}}\\
\small{\text{
$=(1+50) *50 = 2550
$
}}$$

Part 1 + Part 2 = 2550 + 2500x

Hi Melody,

my calculation with your picture of the triangle without trigonometry: