Suppose a sequence $y_n\to y > 0$ and let $a$ be a real number. Determine whether the limit $\lim\limits_{n\to\infty} 2^n \arcsin (\frac{a}{2^n y_n})$ exists (is finite) and if so, determine its value.
I think the answer is $\frac{a}{y},$ but I'm not sure which limit properties to use. Would the Squeeze theorem be useful, for instance? Intuitively, as $n\to\infty, \frac{a}{2^n y_n}\to 0,$ so $\arcsin (\frac{a}{2^n y_n} )\approx \frac{a}{2^ny_n}.$ I'm not sure if I can really apply the product property of limits; i.e. $\lim\limits_{n\to\infty} a_n b_n = \lim\limits_{n\to\infty} a_n\cdot \lim\limits_{n\to\infty} b_n$ provided the RHS value is finite. However, I'm not really sure how to formalize this using the definition of limits. Let $\epsilon > 0.$ I want to show that there exists $N\in\mathbb{N}$ so that for $n\ge N, |2^n \arcsin (\frac{a}{2^n y_n}) -L| < \epsilon,$ where $L$ is the limit of the sequence.