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Here is my problem: $$\begin{align} &\lim_{\beta\to 0^-} \left(-\ln|\beta| + e^{\beta}\right) + \lim_{\beta\to 0^+} \left(\ln|\beta|\right) \\ =& \lim_{\beta\to 0} \left(\ln|\beta| - \ln|\beta| + e^{\beta}\right) \\ =& \lim_{\beta\to 0} e^{\beta} = 1 \end{align}$$

First of all sorry i can't give the whole context since will consume a lot of time to type it by hands. So, in short i'm evaluating the integral using cauchy principal value and came up with this.

Is it legal to do that limit? Since i have an absolute value and if i split the first limit, the first term will be the same as the second limit and for convenience i change the limit goes to 0 (both sides)? Please help me to understand about limit. Or maybe if it's illegal you can tell me why. Thanks in advance!

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  • $\begingroup$ The first equality is valid if both those limits in the first line exist. $\endgroup$
    – tangentbundle
    Jan 27, 2021 at 1:52

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Unfortunately, this is not legal. Why? Well - when you say something like $$ \lim_{\beta\to 0}\ln|\beta| $$ we are talking about the number that this limit converges to. If it doesn't converge, it's not well defined. The property of limits that $$ \lim_{x\to a}f(x)+g(x) = \lim_{x\to a}f(x) + \lim_{x\to a}g(x) $$ (which is essentially what you used on the first line, or I guess, the reverse of this) only holds if the individual limits exist (in your case, they definitely don't).

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    $\begingroup$ Very nice. Thank you so much for your answer! $\endgroup$
    – user516076
    Jan 27, 2021 at 2:01
  • $\begingroup$ @user516076 no problem! $\endgroup$
    – Riemann'sPointyNose
    Jan 27, 2021 at 2:07
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    $\begingroup$ @user516076: Since you can't see my comment on the deleted wrong answer, let me repeat it here for your reference: as Riemann'sPointyNose stated, your limit expressions refer to the actual limits if they exist, but they do not. So your very first expression is undefined, which implies that you did something wrong earlier (so the error is actually hiding in the context you did not want to provide). $\endgroup$
    – user21820
    Jan 27, 2021 at 16:59

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