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anonymous
 4 years ago
lim x>0 ((x+deltaX)^2+x+deltaX(x^2+x))/deltaX
anonymous
 4 years ago
lim x>0 ((x+deltaX)^2+x+deltaX(x^2+x))/deltaX

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0} ((x+\Delta x)^2 +x + \Delta x(x^2+x)/\Delta x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0} \frac {(x+\Delta x)^2 +x + \Delta x(x^2+x)}{\Delta x}\]Your gonna have to simplify that out, although it gets annoying...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{\Delta x \rightarrow 0} \frac {(x+\Delta x)^2 +x + \Delta x(x^2+x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x  x^2  x}{\Delta x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea i did that my problem is after that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i cancel out the x^2 and X

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{\Delta x \rightarrow 0} \frac { x^2 + 2x \Delta x + \Delta^2 x + x + \Delta x  x^2  x}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac { 2x \Delta x + \Delta x + \Delta^2 x}{\Delta x} = \lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02xdeltax+deltax^2+deltaX/deltax

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{\Delta x \rightarrow 0} 2x + 1 + \Delta x = 2x + 1 + 0 = 2x + 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, divide the delta x's out to simplifiy. Then just evaluate the limit by plugging in 0 for delta x.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so i factor the deltax out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0deltax(2x+deltax+1)/1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops i mean over delta x not 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i end up with 2x+deltax+1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0however the book gives me an answer of 1/x^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so im doing something wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so im confused beyond belief atm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01/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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the instruction is to find the limit if it exist, if it does not explain why

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how do they get a limit of 1/x^2 out of 2x+1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well the book does not show me the work in the back, just the answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0guess this is one to ask the instructor :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, 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.
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