limit of sequence involving arcsin

Question

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.

Answer 1

Take $2^{n}y_{n}=ax_{n}$

Then you have $$\lim_{n\to\infty}\frac{ax_{n}}{y_{n}}\arcsin(\frac{1}{x^{n}})$$.

(such a substitution is justified as your question assumes $y_{n}\neq 0$ as it was in the denominator)

Which you can write as $$\lim_{n\to\infty}\frac{a}{y_{n}}x_{n}\arcsin(\frac{1}{x_{n}})$$.

Now you take $a_{n}=\frac{a}{y_{n}}$ and $b_{n}=x_{n}\arcsin(\frac{1}{x_{n}})$.

Now you can use product as both these sequences converge separately. Now $x\cdot\arcsin(\frac{1}{x})$ is a continuous bounded function in $[1,\infty)$. And differentiable in $(1,\infty)$. So you can use that to justify $\lim_{n\to\infty}x_{n}\arcsin(\frac{1}{x_{n}})=\lim_{x\to\infty}x\cdot\arcsin(\frac{1}{x})=\lim_{n\to\infty}\frac{\arcsin(\frac{1}{x})}{\frac{1}{x}}$. Which would justify the use of L'Hospital (as both $\arcsin(\frac{1}{x})$ and $\frac{1}{x}$ are continuous in $[1,\infty)$ and differentiable in $(1,\infty)$.) and concluding that this limit is indeed $1$.

And $\lim_{n\to\infty}a_{n}=\frac{a}{y}$.

So you have your answer as $$\frac{a}{y}$$