Check my proof for this limit relation please
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Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
$endgroup$
add a comment |
$begingroup$
Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
$endgroup$
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
add a comment |
$begingroup$
Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
$endgroup$
Consider two functions $f, g:mathbb{R} rightarrow mathbb{R} $ and $ain mathbb{R} $. We know that $lim_{x to a} f(x) =0$ and $g$ is bounded.
Is it true that $lim_{x to a} f(x) cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $exists M>0$ so that $-Mle g(x) le M$.
If we multiply this relation by
$f(x) $ and apply the squeeze theorem we get that $lim_{x to a} f(x) cdot g(x) =0$.
real-analysis limits proof-verification
real-analysis limits proof-verification
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
asked Jan 8 at 14:55
JustAnAmateurJustAnAmateur
565
565
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
add a comment |
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
$begingroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) le f(x) g(x) le M f(x)$ because $f(x)$ could have been negative.
But yup, if we have $|g(x)| le M$ and we can multiply by $|f(x)|$ and obtain $$0le|f(x)g(x)|le M|f(x)|$$ and now we can use squeeze theorem.
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 8 at 15:05
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
$begingroup$
Thank you! I had had a feeling that I had a silly in my solution, but I am happy that the conclusion is still true.
$endgroup$
– JustAnAmateur
Jan 8 at 15:09
add a comment |
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$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01