# limit of sequence involving arcsin

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.

• Are you allowed to assume knowledge of the limit $\lim_{x\to 0}\frac{\arcsin(x)}{x}=1$ in your solution? Oct 18, 2021 at 3:11

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 $$\frac{a}{y}$$
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