Proof check: limit of a product of two functions $f(x)g(x)$
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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
asked Jan 17 at 20:07
tmarictmaric
20619
$endgroup$
add a comment |
$begingroup$
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 $lim_{x to a} f(x)g(x) = AB$.
calculus limits
asked Jan 17 at 20:07
tmarictmaric
20619
$endgroup$
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
add a comment |
$begingroup$
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 $lim_{x to a} f(x)g(x) = AB$.
calculus limits
asked Jan 17 at 20:07
tmarictmaric
20619
$endgroup$
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
calculus limits
asked Jan 17 at 20:07
tmarictmaric
20619
asked Jan 17 at 20:07
tmarictmaric
20619
edited Jan 18 at 13:21
tmaric
asked Jan 17 at 20:07
tmarictmaric
20619
asked Jan 17 at 20:07
tmarictmaric
20619
asked Jan 17 at 20:07
tmarictmaric
20619
20619
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
add a comment |
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
add a comment |
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$begingroup$
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