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Suppose we want to find $\lim_{x \to a}(f(x) - g(x))$ . We are not aware about the existence of $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ . Can we use sum rule and rewrite it to $\lim_{x \to a}f(x) - \lim_{x \to a}g(x)$ ? For example if after rewriting it , we found that $\lim_{x \to a} f(x) = \infty $ and $\lim_{x \to a} g(x) = l$ then this conclusion $\lim_{x \to a}(f(x) - g(x)) = \infty$ is right ? I always have trouble in the using sum and multiplication rules in order to computing limits .

S.H.W
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  • For the sum rule you need one of the limits to exist (finitely). For product rule you need one of the limits to be non-zero(and finite). This non-zero limit in case of product rule is necessary. For more details see this answer https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Jun 27 '17 at 09:39

1 Answers1

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Just if there is $\lim\limits_{x\rightarrow a}f(x)=A$ and there is $\lim\limits_{x\rightarrow a}g(x)=B$ then there is $\lim\limits_{x\rightarrow a}(f(x)+g(x))=A+B$

About the second question.

If there is $\lim\limits_{x\rightarrow a}f(x)=A$ and $\lim\limits_{x\rightarrow a}g(x)=\infty$ then $\lim\limits_{x\rightarrow a}(f(x)+g(x))=\infty.$

Michael Rozenberg
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  • So the conclusion in my example is wrong ? – S.H.W Jun 26 '17 at 08:19
  • Okay , I know it seems obvious but how we can prove it ? – S.H.W Jun 26 '17 at 08:21
  • Yes , I saw it . Thanks . Please take a look at the second comment . – S.H.W Jun 26 '17 at 08:22
  • @S.H.W We can prove it by the limit definition . I am ready to show if you wish, – Michael Rozenberg Jun 26 '17 at 08:22
  • So , it doesn't relate to the sum rule ? Because I always thought it comes from the sum rule . I'll appreciate that if it's possible . – S.H.W Jun 26 '17 at 08:24
  • @S.H.W Yes, of course! But for the case like $\lim\limits_{x\rightarrow a}f(x)=+\infty$ and $\lim\limits_{x\rightarrow a}g(x)=-\infty$ we can not. In this case we need something more because $ \lim\limits_{x\rightarrow a}(f(x)+g(x))$ can be all number or does not exist. – Michael Rozenberg Jun 26 '17 at 08:27
  • Yes , I see . It depends on the two functions . – S.H.W Jun 26 '17 at 08:29
  • @S.H.W Yes, exactly! – Michael Rozenberg Jun 26 '17 at 08:30
  • Honestly , this problem is related to the previous question that you have answered it . In this article (https://math.berkeley.edu/~peyam/Math1AFa10/Slant%20asymptotes.pdf) is explained about the oblique asymptote . In page 3 , I saw this : $\lim_{x \to \infty} (\frac{f(x)}{x} - a ) = 0 $ then $\lim_{x \to \infty} \frac{f(x)}{x} = a$ . This expression leads me to asking the second question . – S.H.W Jun 26 '17 at 08:41
  • We are not aware about $\lim_{x \to \infty} \frac{f(x)}{x}$ . So how that conclusion is true ? – S.H.W Jun 26 '17 at 08:47
  • @S.H.W just to calculate a limit. If $\lim\limits_{x\rightarrow+\infty}\frac{f(x)}{x}$ does not exist so we have no an asymptote for $x\rightarrow+\infty$. – Michael Rozenberg Jun 26 '17 at 10:11
  • So , it happens because $a$ is a consonant ? – S.H.W Jun 26 '17 at 11:07
  • @S.H.W I did not say that it happens. If it happens then we can try to calculate $\lim\limits_{x\rightarrow a}(f(x)-ax)$. – Michael Rozenberg Jun 26 '17 at 11:22
  • Okay , Thank you a lot . – S.H.W Jun 26 '17 at 11:24