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ChrisV
lim x>0- ((x+deltaX)^2+x+deltaX-(x^2+x))/deltaX
\[\lim_{x \rightarrow 0-} ((x+\Delta x)^2 +x + \Delta x-(x^2+x)/\Delta x\]
\[\lim_{x \rightarrow 0-} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x}\]Your gonna have to simplify that out, although it gets annoying...
\[\lim_{\Delta x \rightarrow 0} \frac {(x+\Delta x)^2 +x + \Delta x-(x^2+x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x - x^2 - x}{\Delta x}\]
yea i did that my problem is after that
i cancel out the x^2 and X
\[\lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x + \Delta x - x^2 - x}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { 2x \Delta x + \Delta x + \Delta^2 x}{\Delta x} = \lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x\]
2xdeltax+deltax^2+deltaX/deltax
\[\lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x = 2x + 1 + 0 = 2x + 1\]
Yeah, divide the delta x's out to simplifiy. Then just evaluate the limit by plugging in 0 for delta x.
ok so i factor the deltax out
oops i mean over delta x not 1
so i end up with 2x+deltax+1
however the book gives me an answer of -1/x^2
so im doing something wrong
so im confused beyond belief atm
-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.
the instruction is to find the limit if it exist, if it does not explain why
so how do they get a limit of 1/x^2 out of 2x+1?
well the book does not show me the work in the back, just the answer
guess this is one to ask the instructor :)
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.