Answer to Question #106692 in Calculus for Parul

Question #106692
a) Obtain the 6th roots of (-7) , represent them in an Argand diagram.
b) using epsilon Delta definition , show that Lim(3x-5)=1
x->2
1
Expert's answer
2020-04-01T14:14:32-0400

a) The 6th root of (7)(-7) are the solution of the equation

z6+7=0z^6+7=0 ............(1)

Where z is a complex number.


z6+7=0     z6=7=7(1)     z=716(1)16z^6+7=0 \ \implies z^6=-7=7(-1) \ \implies z=7^{\frac{1}{6}}(-1)^{\frac{1}{6}}

Again ,we known that

eπi=cos(π)+isin(π)e^{\pi i}=cos(\pi)+i sin(\pi)

=1=-1

Again,e2πki=1e^{2\pi k i}=1 for all kZk\in \Z

Therefore,

eπie2kπi=1     e(2k+1)πi=1e^{\pi i}e^{2k\pi i}=-1 \ \implies e^{(2k+1)\pi i }=-1

for all kZk\in \Z .


z=716e(2k+1)πi6\therefore z= 7^{\frac{1}{6}} e^{\frac{(2k+1)\pi i }{6}}

Where ,k=0,1,2,3,4,5Where \ , k=0,1,2,3,4,5 give the solution of equation (1).

Therefore ,the 6th root of (7)(-7) are


αeπi6,αe3πi6,αe5πi6,αe7πi6,αe9πi6,αe11πi6\alpha e^{\frac{\pi i} {6}},\alpha e^{\frac{3\pi i}{6}},\alpha e^{\frac{5\pi i}{6}}, \alpha e^{\frac{7\pi i}{6}},\alpha e^{\frac{9\pi i}{6}},\alpha e^{\frac{11\pi i}{6}}


Where α=716\alpha=7^{\frac{1}{6}} .

The argand diagram is at following photo.






b) Let f(x)=3x5 , xRf(x)=3x-5 \ ,\forall \ x\in \R

A little algebraic manipulation gives us

f(x)1=3x51=3(x2)=3x2.\mid f(x)-1 \mid=\mid 3x-5-1\mid=\mid 3(x-2)\mid=3\mid x-2\mid.

Now , for a given ϵ>0\epsilon \gt 0 ,we choose δ=ϵ3\delta=\frac{\epsilon}{3} . Then if x2<δ,\mid x-2\mid \lt\delta,

f(x)1=3x51=3x6=3x2<3δ=ϵ\mid f(x)-1\mid=\mid3x-5-1\mid= \mid 3x-6\mid=3\mid x-2\mid \lt3\delta=\epsilon

Since ϵ>0\epsilon \gt0 is arbitrary, we deduce that limx2f(x)=1\lim_{x \to 2}f(x)=1


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