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We have rules for computing limits. One of those is the sum rule: $$\lim\limits_{x\to a}[f(x)+g(x)]=\lim\limits_{x\to a}f(x)+\lim\limits_{x\to a}g(x),\quad \text{provided the limits on the right exist}$$ All rules require the assumption that both $\lim\limits_{x\to a}f(x)$ and $\lim\limits_{x\to a}g(x)$ exist.

So are the following attempts right or wrong ? Although all the results are correct,personally, I feel something wrong (such as it does not make sense to right $\infty +$ a number) but i cannot explain.

$\lim\limits_{x\to \infty} (x+\dfrac{3}{x})=\lim\limits_{x\to \infty} x+\lim\limits_{x\to \infty} \dfrac{3}{x}=\infty +0=\infty $

$\lim\limits_{x\to \infty} (x\ln x+e^x)=\lim\limits_{x\to \infty} x\ln x+\lim\limits_{x\to \infty} e^x=\infty\times\infty+\infty=\infty$

$\lim\limits_{x\to 0} (x\sin \dfrac{1}{x})$

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  • $\begingroup$ The first and second diverges for obvious reason and you could use sandwich theorem for the third one. $\endgroup$
    – cr001
    Nov 2, 2015 at 9:14
  • $\begingroup$ @cr001 ok, so show me the obvious reason. $\endgroup$
    – Bungbu
    Nov 2, 2015 at 9:35
  • $\begingroup$ Well for any number $n$ you state that the function converges to, let $x=n+1$ so the first one is then at least $n+1$ which is strictly larger than $n$. The second one is at least $e^{n+1}$ so obviously strictly larger than $n$. $\endgroup$
    – cr001
    Nov 2, 2015 at 9:44
  • $\begingroup$ @cr001 I really don't get it :( Ok, so let me guess: you are proving by using basic identities of single functions like exponential and log functions, aren't you ? So if you have the sum of 2 more complex functions whose limits are infinity, then how can you determine the limit of the sum of the 2 functions ? $\endgroup$
    – Bungbu
    Nov 2, 2015 at 9:52
  • $\begingroup$ Not really. It is basically just using the fact that ${3\over x}\geq 0$ for any $x$ .etc so $x+{3\over x} \geq x$ for all $x$. If you claim the function converges to $l$ as $x$ go to infinity, then you are saying given any $\epsilon$ there exist an $x_0$ that for all $x>x_0$, $|f(x)-l|$ is less than epsilon, but however say the epsilon is $0.5$ and we let $x_1=l+1$ where $l$ is your proposed limit, then everything larger than $x_1$ we have $|f(x)-l|$ larger than the epsilon, which is obviously a contradiction. $\endgroup$
    – cr001
    Nov 2, 2015 at 9:59

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I think that you have a little bit of a misunderstanding of the rules of evaluating limits; what the sum rule says is IF the individual limits exists, THEN the limit of the sum is the sum of the limits. Notice that theorem says nothing in the other direction. It is perfectly possible to have the limit of the sum of two functions exists without either of the individual limits existing; simply consider $f(x)=\sin(x)$ and $g(x)=-\sin(x)$. Their sum is identically $0$, but each diverges by oscillation by themselves!

Now, as for your given examples, the other answer has provided solutions on how to evaluate them. I personally believe that you should take the content of this answer to heart though; Just because you can use the sum rule if the individual limits exist, doesn't mean that's the ONLY way to evaluate the limit of a sum.

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  • $\begingroup$ Thank you for your answer. I have just editted my question in order to make it clearer. We all can see that the first two limits are divergent and we cannot use sum rule or product rule in these cases. So how can we get the $\infty$ in the limits ? $\endgroup$
    – Bungbu
    Nov 2, 2015 at 9:31

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