# Inconsistency of limits

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?

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.
• For the record, Mathematica gives me a limit of 0, while the same Mathematica query entered into Wolfram Alpha (while parsed correctly) gives me $-\infty$. – greenbagels Apr 5 at 5:03
• @greenbagels ahhh, I actually used Mathematica before Wolfram Alpha but Mathematica still gave me $-\infty$. Hmm........ – Chee Han Apr 5 at 6:40