Check my proof for this limit relation please












2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    looks good to me
    $endgroup$
    – gt6989b
    Jan 8 at 15:01
















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    looks good to me
    $endgroup$
    – gt6989b
    Jan 8 at 15:01














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 14:55









JustAnAmateurJustAnAmateur

1065




1065












  • $begingroup$
    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




$begingroup$
looks good to me
$endgroup$
– gt6989b
Jan 8 at 15:01










1 Answer
1






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 8 at 15:05









Siong Thye GohSiong Thye Goh

101k1466117




101k1466117












  • $begingroup$
    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




$begingroup$
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


















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