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







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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







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      $begingroup$


      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







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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




      calculus limits functions proof-verification epsilon-delta






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      edited Jan 12 at 17:37









      rtybase

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      asked Jan 12 at 17:26









      ArielKArielK

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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}






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          • $begingroup$
            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













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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}






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          • $begingroup$
            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


















          3












          $begingroup$

          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}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          3












          3








          3





          $begingroup$

          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}






          share|cite|improve this answer









          $endgroup$



          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}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            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




















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