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i want to calculate the limit of the following expression when 'w' tend to zero. I have used the Limit function, it takes a lot of time for running without any result. could you please help me how to do that? Thanks in advance

  Limit[1 - ((BesselK[m, w]^2)/(BesselK[m - 1, w]*BesselK[m + 1, w])), w -> 0];

p.s: I know the answer is 1/m, but I want to show this.

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    $\begingroup$ The limit is given by $\frac{1}{m}, \text{for}\; m\in\mathbb{R}$. $\endgroup$
    – yarchik
    Commented Feb 19, 2021 at 20:09
  • $\begingroup$ @yarchik Thanks for your response. I know that. but I want to show this $\endgroup$
    – fatemeh kamali
    Commented Feb 19, 2021 at 20:22
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    $\begingroup$ With suitable assumptions we can get this for m>1. In[334]:= ll = Limit[1 - BesselK[m, w]^2/(BesselK[-1 + m, w] BesselK[1 + m, w]), w -> 0, Direction -> "FromAbove", Assumptions -> m > 1] Out[334]= 1 - Gamma[m]^2/(Gamma[-1 + m] Gamma[1 + m]) In[335]:= Simplify[FunctionExpand[ll, Assumptions -> m > 1]] Out[335]= 1/m $\endgroup$
    – Daniel Lichtblau
    Commented Feb 19, 2021 at 22:03
  • $\begingroup$ @Daniel Lichtblau Thanks a lot! $\endgroup$
    – fatemeh kamali
    Commented Feb 19, 2021 at 23:23

1 Answer 1

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This limit is hard for m Integer...

The following takes about 30 secs to evaluate:

Limit[Limit[Normal[Series[1 - ((BesselK[m, w]^2)/(BesselK[m - 1, w]*
       BesselK[m + 1, w])), {w, 0, 0}]] /. m -> m - e, w -> 0, 
Direction -> "FromAbove"], e -> 0] // FullSimplify

and gives 1/m. But this is only correct for m>=1. For 0<=m<=1 the limit seems to be 1.

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  • $\begingroup$ Thanks a lot for your answer $\endgroup$
    – fatemeh kamali
    Commented Feb 19, 2021 at 23:24

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