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Is there any limit like this

$$\lim_{f(x)\to0} g(x)=0?$$

Defined as follows$$\forall \epsilon>0,\exists \delta>0 | |g(x)|<\epsilon ,\forall x |0<|f(x)|<\delta?$$ Where f and g real functions, such that range of f may not contain $0$.

In the real analysis book by bartle page - 201 (PDF page-217), in the remark following Riemann integral part , it says that the Riemann integral as defined in the text , is often called limit tends to infinity , but the author mentions this limit is not the usual limit as discussed earlier in the book.

So my question is can we define the limit more generally as in above and what are the properties(example additive,multiplicative etc) of this limit, that makes it different from normal limit (where informally the g(x) is an explicit function of the tending variable) ?


Are the following statements true or false? (all the functions are assumed to be real functions and the limits are assumed to exist)

$$\lim_{f(x)\to0} (g(x)+p(x))=\lim_{f(x)\to0} g(x)+\lim_{f(x)\to0} p(x)$$ $$\lim_{f(x)\to0} (g(x)\cdot p(x))=\lim_{f(x)\to0} g(x)\cdot\lim_{f(x)\to0} p(x)$$ Is squeeze theorem valid for this limit?

$$\lim_{g(x)\to0} f(x)+\lim_{p(x)\to0}f(x)=?$$ (is there any formula for the above limit sum?)


Any other theorem concerning the limit is always welcomed.

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    $\begingroup$ $|g(x)|<0$, are you sure ??? $\endgroup$ – Yves Daoust Mar 15 at 14:34
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    $\begingroup$ This is an interesting question. Please write the intended formula in the fourth line of your text correctly. $\endgroup$ – Christian Blatter Mar 15 at 14:36
  • $\begingroup$ @YvesDaoust I think so because the definition of normal limits goes like this.Being inexperienced if I am wrong please correct me. $\endgroup$ – Bijayan Ray Mar 15 at 14:44
  • $\begingroup$ Hem, can an absolute value really be negative ? I am pointing to a typo. $\endgroup$ – Yves Daoust Mar 15 at 14:49
  • $\begingroup$ oh sorry got it , is it now good? $\endgroup$ – Bijayan Ray Mar 15 at 14:53

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