Question

lim x>0- ((x+deltaX)^2+x+deltaX-(x^2+x))/deltaX

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

yea i did that my problem is after that

Answer 5

i cancel out the x^2 and X

Answer 6

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

Answer 7

2xdeltax+deltax^2+deltaX/deltax

Answer 8

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

Answer 9

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

Answer 10

ok so i factor the deltax out

Answer 11

deltax(2x+deltax+1)/1

Answer 12

oops i mean over delta x not 1

Answer 13

so i end up with 2x+deltax+1

Answer 14

set delta x to 0

Answer 15

gives me 2x+1

Answer 16

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

Answer 17

so im doing something wrong

Answer 18

so im confused beyond belief atm

Answer 19

-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.

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

guess this is one to ask the instructor :)

Answer 24

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.

Answer 25

k thanks