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

1. ChrisV Group Title

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

2. Rogue Group Title

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

3. Rogue Group Title

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

4. ChrisV Group Title

yea i did that my problem is after that

5. ChrisV Group Title

i cancel out the x^2 and X

6. Rogue Group Title

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

7. ChrisV Group Title

2xdeltax+deltax^2+deltaX/deltax

8. Rogue Group Title

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

9. Rogue Group Title

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

10. ChrisV Group Title

ok so i factor the deltax out

11. ChrisV Group Title

deltax(2x+deltax+1)/1

12. ChrisV Group Title

oops i mean over delta x not 1

13. ChrisV Group Title

so i end up with 2x+deltax+1

14. ChrisV Group Title

set delta x to 0

15. ChrisV Group Title

gives me 2x+1

16. ChrisV Group Title

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

17. ChrisV Group Title

so im doing something wrong

18. ChrisV Group Title

so im confused beyond belief atm

19. Rogue Group Title

-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 Group Title

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

21. ChrisV Group Title

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

22. ChrisV Group Title

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

23. ChrisV Group Title

guess this is one to ask the instructor :)

24. Rogue Group Title

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 Group Title

k thanks