anonymous
  • anonymous
I NEED HELP HERE
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino PLEASE POST IT HERE WHEN YOU ARE THROUGH
anonymous
  • anonymous
anonymous
  • anonymous
CAN ANYONE HELP PLEASE

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Jack1
  • Jack1
cant open the file... u have a screenshot dude?
anonymous
  • anonymous
no
anonymous
  • anonymous
http://prntscr.com/83uvhy
anonymous
  • anonymous
Screenie for @jack1
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
http://prntscr.com/83uvhy
anonymous
  • anonymous
@Michele_Laino please help with this open question
Michele_Laino
  • Michele_Laino
by definition of the limit value, we can write this: \[\Large \forall \varepsilon ,\exists \delta |{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon \]
Michele_Laino
  • Michele_Laino
that is the definition of the limit value of f(x) when x--->x_0
Michele_Laino
  • Michele_Laino
similarly for g(x): \[\Large \forall \varepsilon ,\exists \delta \;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon \]
Michele_Laino
  • Michele_Laino
where the symbol: \[\Large {\left\| x -x_0 \right\|_{{\mathbb{R}^n}}}\] stands for the norm of R^n
Michele_Laino
  • Michele_Laino
of course I mean \[\Large \forall \varepsilon > 0,\exists \delta > 0\]
Michele_Laino
  • Michele_Laino
namely, \epsilon and \delta are both positive numbers
Michele_Laino
  • Michele_Laino
tell me when I may continue
anonymous
  • anonymous
ok continue sir
Michele_Laino
  • Michele_Laino
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:
Michele_Laino
  • Michele_Laino
by definition, of limit value, we can write: \[\Large \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /\left| a \right|\]
Michele_Laino
  • Michele_Laino
and therefore, we have: \[\large \left| {\left( {af} \right)\left( x \right) - aL} \right| = \left| {af\left( x \right) - aL} \right| = \left| a \right|\left| {f\left( x \right) - L} \right| \leqslant \left| a \right| \cdot \left( {\varepsilon /\left| a \right|} \right) = \varepsilon \]
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} \left| {\left( {af} \right)\left( x \right) - aL} \right| = \left| {af\left( x \right) - aL} \right| = \hfill \\ \hfill \\ = \left| a \right|\left| {f\left( x \right) - L} \right| \leqslant \left| a \right| \cdot \left( {\varepsilon /\left| a \right|} \right) = \varepsilon \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so we are done!
Michele_Laino
  • Michele_Laino
similarly, if g(x)--->M, when x--->x_0, then, we have: (bg)(x)---> bM, when x--->x_0
anonymous
  • anonymous
but we have addition, subtraction, multiplication and division?
Michele_Laino
  • Michele_Laino
yes! I know, at the moment I have proven that, if: f--->L then af--->aL
Michele_Laino
  • Michele_Laino
next I will prove that if f--->L and g--->M, then: f+g--->L+M
anonymous
  • anonymous
ok sir . so is that for 3I in the question?
Michele_Laino
  • Michele_Laino
yes! exactly
Michele_Laino
  • Michele_Laino
here is the proof:
Michele_Laino
  • Michele_Laino
by hypothesis, we can write this: \[\Large \begin{gathered} \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2 \hfill \\ \hfill \\ \forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2 \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
then we have: \[\Large \begin{gathered} \left| {f\left( x \right) + g\left( x \right) - \left( {L + M} \right)} \right| = \left| {f\left( x \right) - L + g\left( x \right) - M} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left| {f\left( x \right) - L} \right| + \left| {g\left( x \right) - M} \right| \leqslant \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so, we are done!
Michele_Laino
  • Michele_Laino
now, the property i) is completely proven
anonymous
  • anonymous
ok sir
Michele_Laino
  • Michele_Laino
next I give the proof of property ii)
Michele_Laino
  • Michele_Laino
here, by hypothesis, we can write this: \[\large \begin{gathered} \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ \forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ K = \max \left( {1 + \left| M \right|,1 + \left| L \right|} \right) \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
herefore, we have the subsequent steps: \[\large \begin{gathered} \left| {f\left( x \right)g\left( x \right) - LM} \right| = \left| {f\left( x \right)g\left( x \right) - Lg\left( x \right) + Lg\left( x \right) - LM} \right| = \hfill \\ \hfill \\ = \left| {\left( {f\left( x \right) - L} \right)g\left( x \right) + L\left( {g\left( x \right) - M} \right)} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left| {g\left( x \right)} \right| \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
I have added and subtracted the quantity L*g(x)
Michele_Laino
  • Michele_Laino
next we note that: \[\left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant 1 + \left| M \right|\]
Michele_Laino
  • Michele_Laino
\[\large \left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant 1 + \left| M \right|\]
Michele_Laino
  • Michele_Laino
more precisely: \[\Large \begin{gathered} \left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant \hfill \\ \hfill \\ \leqslant \varepsilon + \left| M \right| \leqslant 1 + \left| M \right| \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so, we can write: \[\large \begin{gathered} \left| {f\left( x \right)g\left( x \right) - LM} \right| = \left| {f\left( x \right)g\left( x \right) - Lg\left( x \right) + Lg\left( x \right) - LM} \right| = \hfill \\ \hfill \\ = \left| {\left( {f\left( x \right) - L} \right)g\left( x \right) + L\left( {g\left( x \right) - M} \right)} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left| {g\left( x \right)} \right| \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left( {1 + \left| M \right|} \right) \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\left| {f\left( x \right) - L} \right| + \left| {g\left( x \right) - M} \right|} \right) \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\frac{\varepsilon }{{2K}} + \frac{\varepsilon }{{2K}}} \right) = \varepsilon \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so, we are done
anonymous
  • anonymous
thanks. great. so we are left with one now
Michele_Laino
  • Michele_Laino
yes! we have to prove the property iii)
Michele_Laino
  • Michele_Laino
again, by hypothesis, we can write this: \[\large \begin{gathered} \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ \forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ K = \max \left( {\frac{{\left| M \right|}}{{\left| {M - \varepsilon } \right|\left| M \right|}},\frac{{\left| L \right|}}{{\left| {M - \varepsilon } \right|\left| M \right|}}} \right) \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
then we have: \[\large \begin{gathered} \left| {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{L}{M}} \right| = \left| {\frac{{f\left( x \right)M - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \left| {\frac{{f\left( x \right)M - LM + LM - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \frac{{M\left( {f\left( x \right) - L} \right) + L\left( {M - g\left( x \right)} \right)}}{{g\left( x \right)M}} \leqslant \hfill \\ \hfill \\ \leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {g\left( x \right)} \right|\left| M \right|}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
I have added and subtracted the quantity L*M
Michele_Laino
  • Michele_Laino
now, by hypothesis, we can write: \[\Large g\left( x \right) \geqslant M - \frac{\varepsilon }{{2K}} \geqslant M - 1\] or: \[\Large \frac{1}{{g\left( x \right)}} \leqslant \frac{1}{{M - 1}} \Rightarrow \frac{1}{{\left| {g\left( x \right)} \right|}} \leqslant \frac{1}{{\left| {M - 1} \right|}}\]
Michele_Laino
  • Michele_Laino
therefore, we have: \[\large \begin{gathered} \left| {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{L}{M}} \right| = \left| {\frac{{f\left( x \right)M - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \left| {\frac{{f\left( x \right)M - LM + LM - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \frac{{M\left( {f\left( x \right) - L} \right) + L\left( {M - g\left( x \right)} \right)}}{{g\left( x \right)M}} \leqslant \hfill \\ \hfill \\ \leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {g\left( x \right)} \right|\left| M \right|}} \leqslant \hfill \\ \hfill \\ \leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {M - 1} \right|\left| M \right|}} \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\left| {f\left( x \right) - L} \right| + \left| {M - g\left( x \right)} \right|} \right) \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\frac{\varepsilon }{{2K}} + \frac{\varepsilon }{{2K}}} \right) = \varepsilon \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
oops.. the right definition of K is: \[\Large K = \max \left( {\frac{{\left| M \right|}}{{\left| {M - 1} \right|\left| M \right|}},\frac{{\left| L \right|}}{{\left| {M - 1} \right|\left| M \right|}}} \right)\]
Michele_Laino
  • Michele_Laino
so we are done
anonymous
  • anonymous
ok, thanks . will study it please do not delete
anonymous
  • anonymous
please post number 2b solution steps here sir. thanks @Michele_Laino
Michele_Laino
  • Michele_Laino
please, attach that exercise, using the "Attach File" button
Michele_Laino
  • Michele_Laino
more precisely in the last proof, we can suppose, more simply: \[\Large K = \frac{1}{{\left| {M - 1} \right|}}\]

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