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Let $I_n(x)$ and $L_n(x)$ be the modified Bessel and modified Struve functions of order $n$, respectively. Assuming $x$ is real, I am interested in the following limit: $$ \lim_{x\to\infty} \frac{I_0(x)L_1(x) - I_1(x)L_0(x)}{x^2I_2(x)}. $$ Let's call the function $G(x)/x^2$. Now, using Wolfram Alpha I find that $$ \lim_{x\to\infty} G(x) = \lim_{x\to\infty} \frac{I_0(x)L_1(x) - I_1(x)L_0(x)}{I_2(x)} = -\frac{2}{\pi}. $$ So it seems like $$ \lim_{x\to\infty} \frac{I_0(x)L_1(x) - I_1(x)L_0(x)}{x^2I_2(x)} = \lim_{x\to\infty} \frac{G(x)}{x^2} = \left(\lim_{x\to\infty} G(x)\right)\left(\lim_{x\to\infty} \frac{1}{x^2}\right) = 0 $$ On the other hand, Wolfram Alpha gives me
$$ \lim_{x\to\infty} \frac{G(x)}{x^2} = -\infty. $$ What went wrong?

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Ok, I managed to show (using squeeze theorem) that $\displaystyle\lim_{x\to\infty} G(x)/x^2 = 0$. Nonetheless I am still interested in knowing why Wolfram Alpha is giving me the wrong limit.

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  • $\begingroup$ For the record, Mathematica gives me a limit of 0, while the same Mathematica query entered into Wolfram Alpha (while parsed correctly) gives me $-\infty$. $\endgroup$ – greenbagels Apr 5 at 5:03
  • $\begingroup$ @greenbagels ahhh, I actually used Mathematica before Wolfram Alpha but Mathematica still gave me $-\infty$. Hmm........ $\endgroup$ – Chee Han Apr 5 at 6:40
  • $\begingroup$ @greenbagels This is what I entered: Limit[(BesselI[0, x]*StruveL[1, x] - BesselI[1, x]*StruveL[0, x])/(x^2*BesselI[2, x]), x -> Infinity] $\endgroup$ – Chee Han Apr 5 at 6:43
  • $\begingroup$ Hmm... odd; pasting your code into Mathematica still gives me 0. $\endgroup$ – greenbagels Apr 5 at 16:25

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