## anonymous one year ago f(x) = 1/(x + 7) Find the limit. lim Δx→0 (f(x + Δx) − f(x))/Δx

1. anonymous

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

2. anonymous

i was trying (x+$Deltax$ + 7)(x+7)

3. anonymous

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

4. anonymous

hmmm how does that help?

5. anonymous

and how exactly did you get to that point?

6. anonymous

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

7. 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.

8. anonymous

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

9. 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

10. anonymous

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

11. anonymous
12. anonymous

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

13. 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 }$

14. 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

15. anonymous

fist must simply it before substitution by zero.

16. anonymous

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

17. anonymous

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

18. dinamix

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

19. anonymous

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

20. LynFran

by finding its common denominator

21. anonymous

@ASAAD123 yepp i got to that point

22. LynFran

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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

@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

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

28. LynFran

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

29. LynFran

30. dinamix

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

31. anonymous

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

@YadielG this what u want right ?

33. 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

34. dinamix

ok u are welcome mate @YadielG