## ChrisV lim x>0- ((x+deltaX)^2+x+deltaX-(x^2+x))/deltaX 2 years ago 2 years ago

1. ChrisV

$\lim_{x \rightarrow 0-} ((x+\Delta x)^2 +x + \Delta x-(x^2+x)/\Delta x$

2. Rogue

$\lim_{x \rightarrow 0-} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x}$Your gonna have to simplify that out, although it gets annoying...

3. Rogue

$\lim_{\Delta x \rightarrow 0} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x - x^2 - x}{\Delta x}$

4. ChrisV

yea i did that my problem is after that

5. ChrisV

i cancel out the x^2 and X

6. Rogue

$\lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x + \Delta x - x^2 - x}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { 2x \Delta x + \Delta x + \Delta^2 x}{\Delta x} = \lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x$

7. ChrisV

2xdeltax+deltax^2+deltaX/deltax

8. Rogue

$\lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x = 2x + 1 + 0 = 2x + 1$

9. Rogue

Yeah, divide the delta x's out to simplifiy. Then just evaluate the limit by plugging in 0 for delta x.

10. ChrisV

ok so i factor the deltax out

11. ChrisV

deltax(2x+deltax+1)/1

12. ChrisV

oops i mean over delta x not 1

13. ChrisV

so i end up with 2x+deltax+1

14. ChrisV

set delta x to 0

15. ChrisV

gives me 2x+1

16. ChrisV

however the book gives me an answer of -1/x^2

17. ChrisV

so im doing something wrong

18. ChrisV

so im confused beyond belief atm

19. Rogue

-1/x^2 is the answer to another question, probably the derivative of 1/x. The answer to the question you posted here is 2x + 1. There might be an error in the book, or you read it wrong.

20. ChrisV

the instruction is to find the limit if it exist, if it does not explain why

21. ChrisV

so how do they get a limit of 1/x^2 out of 2x+1?

22. ChrisV

well the book does not show me the work in the back, just the answer

23. ChrisV

guess this is one to ask the instructor :)

24. Rogue

No, for your question, 2x + 1 is the correct answer. That's all there is to it. You used the limit formula to find the derivative of x^2 + x.$\frac {dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac {f(x + \Delta x) - f(x)}{\Delta x}$ If you used the same formula for the function 1/x, you would get -1/x^2. You are correct, the textbook is wrong. Don't worry too much about it.

25. ChrisV

k thanks