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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ive tried a couple of different ways but i think im just multiplying the top and the bottom by the wrong values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was trying (x+\[Deltax\] + 7)(x+7)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d }{ dx}\frac{ 1 }{ x+7}\]=\[\lim_{\Delta x \rightarrow 0}\frac{ \frac{ 1 }{ x+\Delta x+7 }\frac{ 1 }{ x+7 } }{ \Delta x}\]=\[\lim_{\Delta x \rightarrow 0}\frac{ 1}{ (x+7)(x+\Delta x+7) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm how does that help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and how exactly did you get to that point?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@YadielG are you want to differentiate this f(x) from definition of differentiation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im still new to Calculus 1 so im not too knowledgeable about the terminology and processes yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can find the value of limit, just substitute Delta x=0 in this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well that gives you undefined because of the delta x on the bottom

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0i like working wih h instead of delta x ... \[\lim_{h \rightarrow 0}\frac{ (\frac{ 1 }{ x+7 }+h)(\frac{ 1 }{ x+7 }) }{ h }\]\[\lim_{h \rightarrow 0}\frac{ \frac{ 1+hx+7h1}{ (x+7) } }{ h }\]\[\lim_{h \rightarrow 0}\frac{ h(x+7) }{ (x+7) }*\frac{ 1 }{ h }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0fist must simply it before substitution by zero.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@LynFran how can you pull out h from the bottom of the small fraction like that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ASAAD123 Yes i understand that much but what im really having trouble with is simplifying it

dinamix
 one year ago
Best ResponseYou've already chosen the best response.1u sould the answer f'(0)= 1/49

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dinamix yeah it's easy if i just take the derivative but i need to do it using that limit equation

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0by finding its common denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ASAAD123 yepp i got to that point

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0shouldnt it be....??dw:1440883037837:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440876093495:dw no (x+h) is considered like one term

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@LynFran no you would replace the x in (1/x+7) with the (x + delta x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@LynFran it is still f(x) but just in terms of (x + delta x) instead of just x

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1@LynFran , \(h=\Delta x\), it's just another representation of the change in a function. You can either write it: \[\lim_{h\rightarrow 0} \frac{f(x+h)  f(x)}{h}\]\[\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)f(x)}{\Delta x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Jhannybean ok i understand that but what i am saying is that you wouldn't replace the x with the original equation

LynFran
 one year ago
Best ResponseYou've already chosen the best response.0really?? consider example f(x)=3x then...dw:1440883501443:dw...ok i got it now much clearly lol

dinamix
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{\Delta x \rightarrow 0} =\frac{ f(0+\Delta x) f(0) }{ \Delta x } = \frac{ 1 }{ (0+7)(0+0+7) }=1/49\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1Don'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.

dinamix
 one year ago
Best ResponseYou've already chosen the best response.1@YadielG this what u want right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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

dinamix
 one year ago
Best ResponseYou've already chosen the best response.1ok u are welcome mate @YadielG
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