▶Let us think following limit $ R $ which is a real number which is finite and non-zero in the region $ 0 < \sigma < 1 $
$$ R=\lim\limits_{ x\to\infty}x^{1-2\sigma}\frac{\frac{\sin(2\sigma)}{x^{1-\sigma}}+1-\left(\frac{x-1}{x}\right)^{1-\sigma}}{\frac{\sin(\sigma)}{x^{\sigma}}+1-\left(\frac{x-1}{x}\right)^{\sigma}} $$
We can not split the right side as two limits. It means:
$$ \lim\limits_{ x\to\infty}x^{1-2\sigma}\cdot \lim\limits_{ x\to\infty}\frac{\frac{\sin(2\sigma)}{x^{1-\sigma}}+1-\left(\frac{x-1}{x}\right)^{1-\sigma}}{\frac{\sin(\sigma)}{x^{\sigma}}+1-\left(\frac{x-1}{x}\right)^{\sigma}} $$
Because:
— The limits in the right side; if one is going to the zero , the other going to the infinity.
–Also, we can not apply L′Hospital rule to the second limit in the right side. Because we can not see the values of $\sin(2\sigma)$ and $\sin(\sigma)$ as zero in the region $0<\sigma<1$.
▶ The question: Am I right, we can not split the right side as two limits ?
Best Answer
You can split a limit as the product of two limits provided both exist and
The case when one is zero and the other one is infinite is the well-known “indeterminate form” $0\cdot\infty$, which exactly means you cannot split the limit into the product of two limits.
I suggest to make the substitution $x=1/t$, which brings the limit to the form $$ \lim_{t\to0^+}t^{2\sigma-1}\frac{t^{1-\sigma}\sin(2\sigma)-(1-t)^{1-\sigma}}{t^\sigma\sin(\sigma)-(1-t)^\sigma} $$ and distinguishing the cases $0<\sigma<1/2$, $\sigma=1/2$ and $1/2<\sigma<1$.