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

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

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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
i was trying (x+\[Deltax\] + 7)(x+7)
\[\frac{ d }{ dx}\frac{ 1 }{ x+7}\]=\[\lim_{\Delta x \rightarrow 0}\frac{ \frac{ 1 }{ x+\Delta x+7 }-\frac{ 1 }{ x+7 } }{ \Delta x}\]=\[\lim_{\Delta x \rightarrow 0}\frac{ -1}{ (x+7)(x+\Delta x+7) }\]

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hmmm how does that help?
and how exactly did you get to that point?
@YadielG are you want to differentiate this f(x) from definition of differentiation?
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.
Im still new to Calculus 1 so im not too knowledgeable about the terminology and processes yet
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
you can find the value of limit, just substitute Delta x=0 in this
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well that gives you undefined because of the delta x on the bottom
i like working wih h instead of delta x ... \[\lim_{h \rightarrow 0}\frac{ (\frac{ 1 }{ x+7 }+h)-(\frac{ 1 }{ x+7 }) }{ h }\]\[\lim_{h \rightarrow 0}\frac{ \frac{ 1+hx+7h-1}{ (x+7) } }{ h }\]\[\lim_{h \rightarrow 0}\frac{ h(x+7) }{ (x+7) }*\frac{ 1 }{ h }\]
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
fist must simply it before substitution by zero.
@LynFran how can you pull out h from the bottom of the small fraction like that?
@ASAAD123 Yes i understand that much but what im really having trouble with is simplifying it
u sould the answer f'(0)= -1/49
@dinamix yeah it's easy if i just take the derivative but i need to do it using that limit equation
by finding its common denominator
@ASAAD123 yepp i got to that point
shouldnt it be....??|dw:1440883037837:dw|
|dw:1440876093495:dw| no (x+h) is considered like one term
@LynFran no you would replace the x in (1/x+7) with the (x + delta x)
@LynFran it is still f(x) but just in terms of (x + delta x) instead of just x
@LynFran , \(h=\Delta x\), it's just another representation of the change in a function. You can either write it: \[\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}\]\[\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\]
@Jhannybean ok i understand that but what i am saying is that you wouldn't replace the x with the original equation
really?? consider example f(x)=3x then...|dw:1440883501443:dw|...ok i got it now much clearly lol
ur right @ASAAD123
\[\lim_{\Delta x \rightarrow 0} =\frac{ f(0+\Delta x) -f(0) }{ \Delta x } = \frac{ -1 }{ (0+7)(0+0+7) }=-1/49\]
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.
@YadielG this what u want right ?
@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
ok u are welcome mate @YadielG

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