Question

I NEED HELP HERE

Answer 1

@Michele_Laino PLEASE POST IT HERE WHEN YOU ARE THROUGH

Answer 2

CAN ANYONE HELP PLEASE

Answer 3

cant open the file... u have a screenshot dude?

Answer 4

no

Answer 5

http://prntscr.com/83uvhy

Answer 6

Screenie for @jack1

Answer 7

@Michele_Laino

Answer 8

http://prntscr.com/83uvhy

Answer 9

@Michele_Laino please help with this open question

Answer 10

by definition of the limit value, we can write this:
\[\Large \forall \varepsilon ,\exists \delta |{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon \]

Answer 11

that is the definition of the limit value of f(x) when x--->x_0

Answer 12

similarly for g(x):
\[\Large \forall \varepsilon ,\exists \delta \;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon \]

Answer 13

where the symbol:
\[\Large {\left\| x -x_0 \right\|_{{\mathbb{R}^n}}}\] stands for the norm of R^n

Answer 14

of course I mean \[\Large \forall \varepsilon > 0,\exists \delta > 0\]

Answer 15

namely, \epsilon and \delta are both positive numbers

Answer 16

tell me when I may continue

Answer 17

ok continue sir

Answer 18

now, we can show, that if f(x)--->L, when x--->x_0, then:
(af)(x)---> aL, when x--->x_0, here is the proof:

Answer 19

by definition, of limit value, we can write:
\[\Large \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /\left| a \right|\]

Answer 20

and therefore, we have:
\[\large \left| {\left( {af} \right)\left( x \right) - aL} \right| = \left| {af\left( x \right) - aL} \right| = \left| a \right|\left| {f\left( x \right) - L} \right| \leqslant \left| a \right| \cdot \left( {\varepsilon /\left| a \right|} \right) = \varepsilon \]

Answer 21

\[\Large \begin{gathered}
\left| {\left( {af} \right)\left( x \right) - aL} \right| = \left| {af\left( x \right) - aL} \right| = \hfill \\
\hfill \\
= \left| a \right|\left| {f\left( x \right) - L} \right| \leqslant \left| a \right| \cdot \left( {\varepsilon /\left| a \right|} \right) = \varepsilon \hfill \\
\end{gathered} \]

Answer 22

so we are done!

Answer 23

similarly, if g(x)--->M, when x--->x_0, then, we have:

(bg)(x)---> bM, when x--->x_0

Answer 24

but we have addition, subtraction, multiplication and division?

Answer 25

yes! I know, at the moment I have proven that, if:
f--->L
then af--->aL

Answer 26

next I will prove that if f--->L and g--->M, then:
f+g--->L+M

Answer 27

ok sir . so is that for 3I in the question?

Answer 28

yes! exactly

Answer 29

here is the proof:

Answer 30

by hypothesis, we can write this:
\[\Large \begin{gathered}
\forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2 \hfill \\
\hfill \\
\forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2 \hfill \\
\end{gathered} \]

Answer 31

then we have:
\[\Large \begin{gathered}
\left| {f\left( x \right) + g\left( x \right) - \left( {L + M} \right)} \right| = \left| {f\left( x \right) - L + g\left( x \right) - M} \right| \leqslant \hfill \\
\hfill \\
\leqslant \left| {f\left( x \right) - L} \right| + \left| {g\left( x \right) - M} \right| \leqslant \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon \hfill \\
\end{gathered} \]

Answer 32

so, we are done!

Answer 33

now, the property i) is completely proven

Answer 34

ok sir

Answer 35

next I give the proof of property ii)

Answer 36

here, by hypothesis, we can write this:
\[\large \begin{gathered}
\forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2K \hfill \\
\hfill \\
\forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2K \hfill \\
\hfill \\
K = \max \left( {1 + \left| M \right|,1 + \left| L \right|} \right) \hfill \\
\end{gathered} \]

Answer 37

herefore, we have the subsequent steps:
\[\large \begin{gathered}
\left| {f\left( x \right)g\left( x \right) - LM} \right| = \left| {f\left( x \right)g\left( x \right) - Lg\left( x \right) + Lg\left( x \right) - LM} \right| = \hfill \\
\hfill \\
= \left| {\left( {f\left( x \right) - L} \right)g\left( x \right) + L\left( {g\left( x \right) - M} \right)} \right| \leqslant \hfill \\
\hfill \\
\leqslant \left| {g\left( x \right)} \right| \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \hfill \\
\end{gathered} \]

Answer 38

I have added and subtracted the quantity L*g(x)

Answer 39

next we note that:
\[\left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant 1 + \left| M \right|\]

Answer 40

\[\large \left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant 1 + \left| M \right|\]

Answer 41

more precisely:
\[\Large \begin{gathered}
\left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant \hfill \\
\hfill \\
\leqslant \varepsilon + \left| M \right| \leqslant 1 + \left| M \right| \hfill \\
\end{gathered} \]

Answer 42

so, we can write:
\[\large \begin{gathered}
\left| {f\left( x \right)g\left( x \right) - LM} \right| = \left| {f\left( x \right)g\left( x \right) - Lg\left( x \right) + Lg\left( x \right) - LM} \right| = \hfill \\
\hfill \\
= \left| {\left( {f\left( x \right) - L} \right)g\left( x \right) + L\left( {g\left( x \right) - M} \right)} \right| \leqslant \hfill \\
\hfill \\
\leqslant \left| {g\left( x \right)} \right| \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \leqslant \hfill \\
\hfill \\
\leqslant \left( {1 + \left| M \right|} \right) \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \leqslant \hfill \\
\hfill \\
\leqslant K\left( {\left| {f\left( x \right) - L} \right| + \left| {g\left( x \right) - M} \right|} \right) \leqslant \hfill \\
\hfill \\
\leqslant K\left( {\frac{\varepsilon }{{2K}} + \frac{\varepsilon }{{2K}}} \right) = \varepsilon \hfill \\
\end{gathered} \]

Answer 43

so, we are done

Answer 44

thanks. great. so we are left with one now

Answer 45

yes! we have to prove the property iii)

Answer 46

again, by hypothesis, we can write this:
\[\large \begin{gathered}
\forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2K \hfill \\
\hfill \\
\forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2K \hfill \\
\hfill \\
K = \max \left( {\frac{{\left| M \right|}}{{\left| {M - \varepsilon } \right|\left| M \right|}},\frac{{\left| L \right|}}{{\left| {M - \varepsilon } \right|\left| M \right|}}} \right) \hfill \\
\end{gathered} \]

Answer 47

then we have:
\[\large \begin{gathered}
\left| {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{L}{M}} \right| = \left| {\frac{{f\left( x \right)M - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\
\hfill \\
= \left| {\frac{{f\left( x \right)M - LM + LM - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\
\hfill \\
= \frac{{M\left( {f\left( x \right) - L} \right) + L\left( {M - g\left( x \right)} \right)}}{{g\left( x \right)M}} \leqslant \hfill \\
\hfill \\
\leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {g\left( x \right)} \right|\left| M \right|}} \hfill \\
\end{gathered} \]

Answer 48

I have added and subtracted the quantity L*M

Answer 49

now, by hypothesis, we can write:
\[\Large g\left( x \right) \geqslant M - \frac{\varepsilon }{{2K}} \geqslant M - 1\]
or:
\[\Large \frac{1}{{g\left( x \right)}} \leqslant \frac{1}{{M - 1}} \Rightarrow \frac{1}{{\left| {g\left( x \right)} \right|}} \leqslant \frac{1}{{\left| {M - 1} \right|}}\]

Answer 50

therefore, we have:
\[\large \begin{gathered}
\left| {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{L}{M}} \right| = \left| {\frac{{f\left( x \right)M - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\
\hfill \\
= \left| {\frac{{f\left( x \right)M - LM + LM - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\
\hfill \\
= \frac{{M\left( {f\left( x \right) - L} \right) + L\left( {M - g\left( x \right)} \right)}}{{g\left( x \right)M}} \leqslant \hfill \\
\hfill \\
\leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {g\left( x \right)} \right|\left| M \right|}} \leqslant \hfill \\
\hfill \\
\leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {M - 1} \right|\left| M \right|}} \leqslant \hfill \\
\hfill \\
\leqslant K\left( {\left| {f\left( x \right) - L} \right| + \left| {M - g\left( x \right)} \right|} \right) \leqslant \hfill \\
\hfill \\
\leqslant K\left( {\frac{\varepsilon }{{2K}} + \frac{\varepsilon }{{2K}}} \right) = \varepsilon \hfill \\
\end{gathered} \]

Answer 51

oops.. the right definition of K is:
\[\Large K = \max \left( {\frac{{\left| M \right|}}{{\left| {M - 1} \right|\left| M \right|}},\frac{{\left| L \right|}}{{\left| {M - 1} \right|\left| M \right|}}} \right)\]

Answer 52

so we are done

Answer 53

ok, thanks . will study it please do not delete

Answer 54

please post number 2b solution steps here sir. thanks @Michele_Laino

Answer 55

please, attach that exercise, using the "Attach File" button

Answer 56

more precisely in the last proof, we can suppose, more simply:
\[\Large K = \frac{1}{{\left| {M - 1} \right|}}\]