Proof check: limit of a product of two functions $f(x)g(x)$












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$begingroup$


If $lim_{x to a}f(x) = A$ and $lim_{x to a}g(x) = B$, $forall epsilon >0$, there is a $delta$ s.t. $0 < |x - a| < delta$ then



$|f(x) - A| < epsilon$ and $|g(x) - a|<epsilon$, (1)



prove that $forall kappa > 0$ there exists $gamma$, s.t. $0 < |x - a| < gamma$ then



$|f(x)g(x) - Acdot B|< kappa$



My attempt at a proof:



$(f(x) - A)(g(x) - B) = f(x)g(x) - Acdot B - B(f(x) - A) - A(g(x) - B)$



$f(x)g(x) - Acdot B = (f(x) - A)(g(x) - B) + B(f(x) - A) + A(g(x) - B)$



$|f(x)g(x) - Acdot B| < |f(x) - A||g(x) - B| + |B||f(x) - A| + |A||g(x) - B|$



Now I can use (1)



$|f(x)g(x) - Acdot B| < epsilon^2 + |B|epsilon + |A|epsilon$



I can now choose $kappa$, such that $kappa = epsilon^2 + |B|epsilon + |A|epsilon$ (this equation has always a positive soluton for $epsilon(kappa)$)., then for this $kappa$, $gamma = delta$ and



$|f(x)g(x) - Acdot B| < kappa$



so $lim_{x to a} f(x)g(x) = AB$.










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  • 1




    $begingroup$
    Be careful in the definition of the neighborhood of $a$ (first few lines). It should be $0<|x-a|<delta$, and same for $gamma$.
    $endgroup$
    – Matteo
    Jan 17 at 20:11








  • 1




    $begingroup$
    Also, in the definition on $lim_{x to a} f(x)$, you have the quantifiers in the wrong order. The quantifier $forall epsilon$ should precede, $exists delta$, i.e. you should write, “$forall epsilon > 0$, $exists delta$ so that if $0 < |x -a| < delta$, then $|f(x) -A| < epsilon$ and $|g(x) - B| < epsilon$.
    $endgroup$
    – Jordan Green
    Jan 17 at 20:15
















0












$begingroup$


If $lim_{x to a}f(x) = A$ and $lim_{x to a}g(x) = B$, $forall epsilon >0$, there is a $delta$ s.t. $0 < |x - a| < delta$ then



$|f(x) - A| < epsilon$ and $|g(x) - a|<epsilon$, (1)



prove that $forall kappa > 0$ there exists $gamma$, s.t. $0 < |x - a| < gamma$ then



$|f(x)g(x) - Acdot B|< kappa$



My attempt at a proof:



$(f(x) - A)(g(x) - B) = f(x)g(x) - Acdot B - B(f(x) - A) - A(g(x) - B)$



$f(x)g(x) - Acdot B = (f(x) - A)(g(x) - B) + B(f(x) - A) + A(g(x) - B)$



$|f(x)g(x) - Acdot B| < |f(x) - A||g(x) - B| + |B||f(x) - A| + |A||g(x) - B|$



Now I can use (1)



$|f(x)g(x) - Acdot B| < epsilon^2 + |B|epsilon + |A|epsilon$



I can now choose $kappa$, such that $kappa = epsilon^2 + |B|epsilon + |A|epsilon$ (this equation has always a positive soluton for $epsilon(kappa)$)., then for this $kappa$, $gamma = delta$ and



$|f(x)g(x) - Acdot B| < kappa$



so $lim_{x to a} f(x)g(x) = AB$.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Be careful in the definition of the neighborhood of $a$ (first few lines). It should be $0<|x-a|<delta$, and same for $gamma$.
    $endgroup$
    – Matteo
    Jan 17 at 20:11








  • 1




    $begingroup$
    Also, in the definition on $lim_{x to a} f(x)$, you have the quantifiers in the wrong order. The quantifier $forall epsilon$ should precede, $exists delta$, i.e. you should write, “$forall epsilon > 0$, $exists delta$ so that if $0 < |x -a| < delta$, then $|f(x) -A| < epsilon$ and $|g(x) - B| < epsilon$.
    $endgroup$
    – Jordan Green
    Jan 17 at 20:15














0












0








0





$begingroup$


If $lim_{x to a}f(x) = A$ and $lim_{x to a}g(x) = B$, $forall epsilon >0$, there is a $delta$ s.t. $0 < |x - a| < delta$ then



$|f(x) - A| < epsilon$ and $|g(x) - a|<epsilon$, (1)



prove that $forall kappa > 0$ there exists $gamma$, s.t. $0 < |x - a| < gamma$ then



$|f(x)g(x) - Acdot B|< kappa$



My attempt at a proof:



$(f(x) - A)(g(x) - B) = f(x)g(x) - Acdot B - B(f(x) - A) - A(g(x) - B)$



$f(x)g(x) - Acdot B = (f(x) - A)(g(x) - B) + B(f(x) - A) + A(g(x) - B)$



$|f(x)g(x) - Acdot B| < |f(x) - A||g(x) - B| + |B||f(x) - A| + |A||g(x) - B|$



Now I can use (1)



$|f(x)g(x) - Acdot B| < epsilon^2 + |B|epsilon + |A|epsilon$



I can now choose $kappa$, such that $kappa = epsilon^2 + |B|epsilon + |A|epsilon$ (this equation has always a positive soluton for $epsilon(kappa)$)., then for this $kappa$, $gamma = delta$ and



$|f(x)g(x) - Acdot B| < kappa$



so $lim_{x to a} f(x)g(x) = AB$.










share|cite|improve this question











$endgroup$




If $lim_{x to a}f(x) = A$ and $lim_{x to a}g(x) = B$, $forall epsilon >0$, there is a $delta$ s.t. $0 < |x - a| < delta$ then



$|f(x) - A| < epsilon$ and $|g(x) - a|<epsilon$, (1)



prove that $forall kappa > 0$ there exists $gamma$, s.t. $0 < |x - a| < gamma$ then



$|f(x)g(x) - Acdot B|< kappa$



My attempt at a proof:



$(f(x) - A)(g(x) - B) = f(x)g(x) - Acdot B - B(f(x) - A) - A(g(x) - B)$



$f(x)g(x) - Acdot B = (f(x) - A)(g(x) - B) + B(f(x) - A) + A(g(x) - B)$



$|f(x)g(x) - Acdot B| < |f(x) - A||g(x) - B| + |B||f(x) - A| + |A||g(x) - B|$



Now I can use (1)



$|f(x)g(x) - Acdot B| < epsilon^2 + |B|epsilon + |A|epsilon$



I can now choose $kappa$, such that $kappa = epsilon^2 + |B|epsilon + |A|epsilon$ (this equation has always a positive soluton for $epsilon(kappa)$)., then for this $kappa$, $gamma = delta$ and



$|f(x)g(x) - Acdot B| < kappa$



so $lim_{x to a} f(x)g(x) = AB$.







calculus limits






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edited Jan 18 at 13:21







tmaric

















asked Jan 17 at 20:07









tmarictmaric

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  • 1




    $begingroup$
    Be careful in the definition of the neighborhood of $a$ (first few lines). It should be $0<|x-a|<delta$, and same for $gamma$.
    $endgroup$
    – Matteo
    Jan 17 at 20:11








  • 1




    $begingroup$
    Also, in the definition on $lim_{x to a} f(x)$, you have the quantifiers in the wrong order. The quantifier $forall epsilon$ should precede, $exists delta$, i.e. you should write, “$forall epsilon > 0$, $exists delta$ so that if $0 < |x -a| < delta$, then $|f(x) -A| < epsilon$ and $|g(x) - B| < epsilon$.
    $endgroup$
    – Jordan Green
    Jan 17 at 20:15














  • 1




    $begingroup$
    Be careful in the definition of the neighborhood of $a$ (first few lines). It should be $0<|x-a|<delta$, and same for $gamma$.
    $endgroup$
    – Matteo
    Jan 17 at 20:11








  • 1




    $begingroup$
    Also, in the definition on $lim_{x to a} f(x)$, you have the quantifiers in the wrong order. The quantifier $forall epsilon$ should precede, $exists delta$, i.e. you should write, “$forall epsilon > 0$, $exists delta$ so that if $0 < |x -a| < delta$, then $|f(x) -A| < epsilon$ and $|g(x) - B| < epsilon$.
    $endgroup$
    – Jordan Green
    Jan 17 at 20:15








1




1




$begingroup$
Be careful in the definition of the neighborhood of $a$ (first few lines). It should be $0<|x-a|<delta$, and same for $gamma$.
$endgroup$
– Matteo
Jan 17 at 20:11






$begingroup$
Be careful in the definition of the neighborhood of $a$ (first few lines). It should be $0<|x-a|<delta$, and same for $gamma$.
$endgroup$
– Matteo
Jan 17 at 20:11






1




1




$begingroup$
Also, in the definition on $lim_{x to a} f(x)$, you have the quantifiers in the wrong order. The quantifier $forall epsilon$ should precede, $exists delta$, i.e. you should write, “$forall epsilon > 0$, $exists delta$ so that if $0 < |x -a| < delta$, then $|f(x) -A| < epsilon$ and $|g(x) - B| < epsilon$.
$endgroup$
– Jordan Green
Jan 17 at 20:15




$begingroup$
Also, in the definition on $lim_{x to a} f(x)$, you have the quantifiers in the wrong order. The quantifier $forall epsilon$ should precede, $exists delta$, i.e. you should write, “$forall epsilon > 0$, $exists delta$ so that if $0 < |x -a| < delta$, then $|f(x) -A| < epsilon$ and $|g(x) - B| < epsilon$.
$endgroup$
– Jordan Green
Jan 17 at 20:15










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