Mixing
Is my proof that $ limlimits_{x to 2} left(x^2 - 3xright) = -2 $ correct?
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My proof: take $x$ such that $|{x - 2}| < 1$. Since $|{x}| -|{2}| < |{x - 2}|$, then $|{x}| < 3$.
Now we want to find a number $n$ such that $|{x - 2}| <1 leq frac{epsilon}{n} $ or $|{x - 2}| < frac{epsilon}{n} leq 1 $ and also so that $|{x - 1}| < n$, and since $|{x}| - |{1}| < |{x - 1}|$ we have that $|{x}| < n + 1$ so we can choose $n = 2$ and by definition of limit, for all $epsilon$ we have at least one $delta$ equal to the smaller between 1 and $frac{epsilon}{2}$ such that $|{x - 2}| < delta$ implies $|{x - 1}||{x - 2}| = |{x^2 - 3x - (-2)}| < frac{delta}{2} = epsilon$.
calculus limits functions proof-verification epsilon-delta
edited Jan 12 at 17:37
rtybase
10.9k21533
asked Jan 12 at 17:26
ArielKArielK
386
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add a comment |
$begingroup$
My proof: take $x$ such that $|{x - 2}| < 1$. Since $|{x}| -|{2}| < |{x - 2}|$, then $|{x}| < 3$.
Now we want to find a number $n$ such that $|{x - 2}| <1 leq frac{epsilon}{n} $ or $|{x - 2}| < frac{epsilon}{n} leq 1 $ and also so that $|{x - 1}| < n$, and since $|{x}| - |{1}| < |{x - 1}|$ we have that $|{x}| < n + 1$ so we can choose $n = 2$ and by definition of limit, for all $epsilon$ we have at least one $delta$ equal to the smaller between 1 and $frac{epsilon}{2}$ such that $|{x - 2}| < delta$ implies $|{x - 1}||{x - 2}| = |{x^2 - 3x - (-2)}| < frac{delta}{2} = epsilon$.
calculus limits functions proof-verification epsilon-delta
edited Jan 12 at 17:37
rtybase
10.9k21533
asked Jan 12 at 17:26
ArielKArielK
386
$endgroup$
add a comment |
$begingroup$
My proof: take $x$ such that $|{x - 2}| < 1$. Since $|{x}| -|{2}| < |{x - 2}|$, then $|{x}| < 3$.
Now we want to find a number $n$ such that $|{x - 2}| <1 leq frac{epsilon}{n} $ or $|{x - 2}| < frac{epsilon}{n} leq 1 $ and also so that $|{x - 1}| < n$, and since $|{x}| - |{1}| < |{x - 1}|$ we have that $|{x}| < n + 1$ so we can choose $n = 2$ and by definition of limit, for all $epsilon$ we have at least one $delta$ equal to the smaller between 1 and $frac{epsilon}{2}$ such that $|{x - 2}| < delta$ implies $|{x - 1}||{x - 2}| = |{x^2 - 3x - (-2)}| < frac{delta}{2} = epsilon$.
calculus limits functions proof-verification epsilon-delta
edited Jan 12 at 17:37
rtybase
10.9k21533
asked Jan 12 at 17:26
ArielKArielK
386
$endgroup$
My proof: take $x$ such that $|{x - 2}| < 1$. Since $|{x}| -|{2}| < |{x - 2}|$, then $|{x}| < 3$.
Now we want to find a number $n$ such that $|{x - 2}| <1 leq frac{epsilon}{n} $ or $|{x - 2}| < frac{epsilon}{n} leq 1 $ and also so that $|{x - 1}| < n$, and since $|{x}| - |{1}| < |{x - 1}|$ we have that $|{x}| < n + 1$ so we can choose $n = 2$ and by definition of limit, for all $epsilon$ we have at least one $delta$ equal to the smaller between 1 and $frac{epsilon}{2}$ such that $|{x - 2}| < delta$ implies $|{x - 1}||{x - 2}| = |{x^2 - 3x - (-2)}| < frac{delta}{2} = epsilon$.
calculus limits functions proof-verification epsilon-delta
calculus limits functions proof-verification epsilon-delta
edited Jan 12 at 17:37
rtybase
10.9k21533
asked Jan 12 at 17:26
ArielKArielK
386
edited Jan 12 at 17:37
rtybase
10.9k21533
asked Jan 12 at 17:26
ArielKArielK
386
edited Jan 12 at 17:37
rtybase
10.9k21533
edited Jan 12 at 17:37
rtybase
10.9k21533
edited Jan 12 at 17:37
rtybase
10.9k21533
10.9k21533
asked Jan 12 at 17:26
ArielKArielK
386
asked Jan 12 at 17:26
ArielKArielK
386
asked Jan 12 at 17:26
ArielKArielK
386
386
add a comment |
add a comment |
1 Answer
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No, it is not correct. In the first place, you start talking about a number $n$ without saying what that number is. And then you state that, by definition of limit, a certain $delta$ must exist. But the goal is to prove that such a $delta$ exists. And how do you know that $1leqslantfracvarepsilon n$?
You can do it as follows: start assuming that $lvert x-2rvert<1$. It follows from this that $lvert x-1rvert<2$. Now, given $varepsilon>0$, take $delta=minleft{fracvarepsilon2,1right}$. Then, since $x^2-3x-(-2)=(x-1)(x-2)$, assuming that $lvert x-2rvert<delta$, you havebegin{align}lvert x^2-3x-(-2)rvert&=bigllvert(x-1)(x-2)bigrrvert\&=lvert x-1rvert.lvert x-2rvert\&<2timesfracvarepsilon2\&=varepsilon.end{align}
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Thank you José :) , that was what I meant actually but I didn't knew how to write it properly in English
$endgroup$
– ArielK
Jan 12 at 17:44
add a comment |
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$begingroup$
No, it is not correct. In the first place, you start talking about a number $n$ without saying what that number is. And then you state that, by definition of limit, a certain $delta$ must exist. But the goal is to prove that such a $delta$ exists. And how do you know that $1leqslantfracvarepsilon n$?
You can do it as follows: start assuming that $lvert x-2rvert<1$. It follows from this that $lvert x-1rvert<2$. Now, given $varepsilon>0$, take $delta=minleft{fracvarepsilon2,1right}$. Then, since $x^2-3x-(-2)=(x-1)(x-2)$, assuming that $lvert x-2rvert<delta$, you havebegin{align}lvert x^2-3x-(-2)rvert&=bigllvert(x-1)(x-2)bigrrvert\&=lvert x-1rvert.lvert x-2rvert\&<2timesfracvarepsilon2\&=varepsilon.end{align}
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Thank you José :) , that was what I meant actually but I didn't knew how to write it properly in English
$endgroup$
– ArielK
Jan 12 at 17:44
add a comment |
$begingroup$
No, it is not correct. In the first place, you start talking about a number $n$ without saying what that number is. And then you state that, by definition of limit, a certain $delta$ must exist. But the goal is to prove that such a $delta$ exists. And how do you know that $1leqslantfracvarepsilon n$?
You can do it as follows: start assuming that $lvert x-2rvert<1$. It follows from this that $lvert x-1rvert<2$. Now, given $varepsilon>0$, take $delta=minleft{fracvarepsilon2,1right}$. Then, since $x^2-3x-(-2)=(x-1)(x-2)$, assuming that $lvert x-2rvert<delta$, you havebegin{align}lvert x^2-3x-(-2)rvert&=bigllvert(x-1)(x-2)bigrrvert\&=lvert x-1rvert.lvert x-2rvert\&<2timesfracvarepsilon2\&=varepsilon.end{align}
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Thank you José :) , that was what I meant actually but I didn't knew how to write it properly in English
$endgroup$
– ArielK
Jan 12 at 17:44
add a comment |
$begingroup$
No, it is not correct. In the first place, you start talking about a number $n$ without saying what that number is. And then you state that, by definition of limit, a certain $delta$ must exist. But the goal is to prove that such a $delta$ exists. And how do you know that $1leqslantfracvarepsilon n$?
You can do it as follows: start assuming that $lvert x-2rvert<1$. It follows from this that $lvert x-1rvert<2$. Now, given $varepsilon>0$, take $delta=minleft{fracvarepsilon2,1right}$. Then, since $x^2-3x-(-2)=(x-1)(x-2)$, assuming that $lvert x-2rvert<delta$, you havebegin{align}lvert x^2-3x-(-2)rvert&=bigllvert(x-1)(x-2)bigrrvert\&=lvert x-1rvert.lvert x-2rvert\&<2timesfracvarepsilon2\&=varepsilon.end{align}
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
No, it is not correct. In the first place, you start talking about a number $n$ without saying what that number is. And then you state that, by definition of limit, a certain $delta$ must exist. But the goal is to prove that such a $delta$ exists. And how do you know that $1leqslantfracvarepsilon n$?
You can do it as follows: start assuming that $lvert x-2rvert<1$. It follows from this that $lvert x-1rvert<2$. Now, given $varepsilon>0$, take $delta=minleft{fracvarepsilon2,1right}$. Then, since $x^2-3x-(-2)=(x-1)(x-2)$, assuming that $lvert x-2rvert<delta$, you havebegin{align}lvert x^2-3x-(-2)rvert&=bigllvert(x-1)(x-2)bigrrvert\&=lvert x-1rvert.lvert x-2rvert\&<2timesfracvarepsilon2\&=varepsilon.end{align}
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 12 at 17:36
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
$begingroup$
Thank you José :) , that was what I meant actually but I didn't knew how to write it properly in English
$endgroup$
– ArielK
Jan 12 at 17:44
add a comment |
$begingroup$
Thank you José :) , that was what I meant actually but I didn't knew how to write it properly in English
$endgroup$
– ArielK
Jan 12 at 17:44
$begingroup$
Thank you José :) , that was what I meant actually but I didn't knew how to write it properly in English
$endgroup$
– ArielK
Jan 12 at 17:44
$begingroup$
Thank you José :) , that was what I meant actually but I didn't knew how to write it properly in English
$endgroup$
– ArielK
Jan 12 at 17:44
add a comment |
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