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

- anonymous

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

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- schrodinger

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- anonymous

Ive tried a couple of different ways but i think im just multiplying the top and the bottom by the wrong values

- anonymous

i was trying (x+\[Deltax\] + 7)(x+7)

- anonymous

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

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## More answers

- anonymous

hmmm how does that help?

- anonymous

and how exactly did you get to that point?

- anonymous

@YadielG are you want to differentiate this f(x) from definition of differentiation?

- anonymous

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.

- anonymous

Im still new to Calculus 1 so im not too knowledgeable about the terminology and processes yet

- anonymous

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

- anonymous

you can find the value of limit, just substitute Delta x=0 in this

- anonymous

http://prntscr.com/8ac5if

- anonymous

well that gives you undefined because of the delta x on the bottom

- LynFran

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

- anonymous

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

- anonymous

fist must simply it before substitution by zero.

- anonymous

@LynFran how can you pull out h from the bottom of the small fraction like that?

- anonymous

@ASAAD123 Yes i understand that much but what im really having trouble with is simplifying it

- dinamix

u sould the answer f'(0)= -1/49

- anonymous

@dinamix yeah it's easy if i just take the derivative but i need to do it using that limit equation

- LynFran

by finding its common denominator

- anonymous

@ASAAD123 yepp i got to that point

- LynFran

shouldnt it be....??|dw:1440883037837:dw|

- anonymous

|dw:1440876093495:dw| no (x+h) is considered like one term

- anonymous

@LynFran no you would replace the x in (1/x+7) with the (x + delta x)

- anonymous

@LynFran it is still f(x) but just in terms of (x + delta x) instead of just x

- Jhannybean

@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

@Jhannybean ok i understand that but what i am saying is that you wouldn't replace the x with the original equation

- LynFran

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

- LynFran

ur right @ASAAD123

- dinamix

\[\lim_{\Delta x \rightarrow 0} =\frac{ f(0+\Delta x) -f(0) }{ \Delta x } = \frac{ -1 }{ (0+7)(0+0+7) }=-1/49\]

- Jhannybean

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.

- dinamix

@YadielG this what u want right ?

- anonymous

@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

ok u are welcome mate @YadielG

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