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anonymous

  • one year ago

f(x) = 1/(x + 7) Find the limit. lim Δx→0 (f(x + Δx) − f(x))/Δx

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  1. anonymous
    • one year ago
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    Ive tried a couple of different ways but i think im just multiplying the top and the bottom by the wrong values

  2. anonymous
    • one year ago
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    i was trying (x+\[Deltax\] + 7)(x+7)

  3. anonymous
    • one year ago
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    \[\frac{ d }{ dx}\frac{ 1 }{ x+7}\]=\[\lim_{\Delta x \rightarrow 0}\frac{ \frac{ 1 }{ x+\Delta x+7 }-\frac{ 1 }{ x+7 } }{ \Delta x}\]=\[\lim_{\Delta x \rightarrow 0}\frac{ -1}{ (x+7)(x+\Delta x+7) }\]

  4. anonymous
    • one year ago
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    hmmm how does that help?

  5. anonymous
    • one year ago
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    and how exactly did you get to that point?

  6. anonymous
    • one year ago
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    @YadielG are you want to differentiate this f(x) from definition of differentiation?

  7. anonymous
    • one year ago
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    I know that the second equation is how you define a derivative using limits, but i just want to find the value of the limit using that second equation.

  8. anonymous
    • one year ago
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    Im still new to Calculus 1 so im not too knowledgeable about the terminology and processes yet

  9. anonymous
    • one year ago
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    i know the equation does have a limit that exists and that it is not zero, but im just having trouble with the algebra getting to that value

  10. anonymous
    • one year ago
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    you can find the value of limit, just substitute Delta x=0 in this

  11. anonymous
    • one year ago
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    http://prntscr.com/8ac5if

  12. anonymous
    • one year ago
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    well that gives you undefined because of the delta x on the bottom

  13. LynFran
    • one year ago
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    i like working wih h instead of delta x ... \[\lim_{h \rightarrow 0}\frac{ (\frac{ 1 }{ x+7 }+h)-(\frac{ 1 }{ x+7 }) }{ h }\]\[\lim_{h \rightarrow 0}\frac{ \frac{ 1+hx+7h-1}{ (x+7) } }{ h }\]\[\lim_{h \rightarrow 0}\frac{ h(x+7) }{ (x+7) }*\frac{ 1 }{ h }\]

  14. anonymous
    • one year ago
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    i thought the goal in this problem was to get rid of all of the x's and just have delta x in order to find the limit

  15. anonymous
    • one year ago
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    fist must simply it before substitution by zero.

  16. anonymous
    • one year ago
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    @LynFran how can you pull out h from the bottom of the small fraction like that?

  17. anonymous
    • one year ago
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    @ASAAD123 Yes i understand that much but what im really having trouble with is simplifying it

  18. dinamix
    • one year ago
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    u sould the answer f'(0)= -1/49

  19. anonymous
    • one year ago
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    @dinamix yeah it's easy if i just take the derivative but i need to do it using that limit equation

  20. LynFran
    • one year ago
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    by finding its common denominator

  21. anonymous
    • one year ago
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    @ASAAD123 yepp i got to that point

  22. LynFran
    • one year ago
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    shouldnt it be....??|dw:1440883037837:dw|

  23. anonymous
    • one year ago
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    |dw:1440876093495:dw| no (x+h) is considered like one term

  24. anonymous
    • one year ago
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    @LynFran no you would replace the x in (1/x+7) with the (x + delta x)

  25. anonymous
    • one year ago
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    @LynFran it is still f(x) but just in terms of (x + delta x) instead of just x

  26. Jhannybean
    • one year ago
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    @LynFran , \(h=\Delta x\), it's just another representation of the change in a function. You can either write it: \[\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}\]\[\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\]

  27. anonymous
    • one year ago
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    @Jhannybean ok i understand that but what i am saying is that you wouldn't replace the x with the original equation

  28. LynFran
    • one year ago
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    really?? consider example f(x)=3x then...|dw:1440883501443:dw|...ok i got it now much clearly lol

  29. LynFran
    • one year ago
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    ur right @ASAAD123

  30. dinamix
    • one year ago
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    \[\lim_{\Delta x \rightarrow 0} =\frac{ f(0+\Delta x) -f(0) }{ \Delta x } = \frac{ -1 }{ (0+7)(0+0+7) }=-1/49\]

  31. Jhannybean
    • one year ago
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    Don't think of x being `replaced`... the concept of limits is in understanding that you're `increasing` the value of x by incriments of \(\Delta x\) or \(h\) as your function converges to 0.

  32. dinamix
    • one year ago
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    @YadielG this what u want right ?

  33. anonymous
    • one year ago
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    @dinamix yes that is exactly what i wanted, interesting though because i was only thinking about replacing delta x with zero instead of x, thank you

  34. dinamix
    • one year ago
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    ok u are welcome mate @YadielG

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