Let $f:R\to R$ and $g:R\to R$ be periodic functions with period $\frac{3}{2}$ and $\frac{1}{2}$ respectively such that $$\lim_{x\to 0}\frac{f(x)}{x}=1$$ and $$\lim_{x\to 0}\frac{g(x)}{x}=2$$
then evaluate the limit $$\lim_{n\to\infty} \frac{f\left(\frac{3}{2}(3+\sqrt{7})^n\right)}{g\left(\frac{1}{2}(2+\sqrt{2})^n\right)}$$
My Attempt
I assumed $f(x)=\frac{3}{4\pi}\sin\left(\frac{4\pi x}{3}\right)$ and $g(x)=\frac{1}{2\pi}\sin(4\pi x)$ and got answer as $0$. Could there be a better way to do it
Best Answer
You can't use specific examples for $f, g$. Instead you need to use the properties of $f, g$ given in question.
Let $$a=(3+\sqrt {7})^n,b=(3-\sqrt {7})^n,c=a+b$$ and $$p=(2+\sqrt {2})^n,q=(2-\sqrt{2})^n,r=p+q$$ so that $a, b, c, p, q, r$ are variables depending on $n$.
Let us note that $c, r$ are positive integers (via binomial theorem) and $b, q$ tend to $0$ as $n\to\infty $. The expression under limit can be written as $$\frac {f(3a/2)}{g(p/2)}=\frac{f(3c/2-3b/2)}{g(r/2-q/2)}$$ and by periodic nature of $f, g$ the above expression equals $$\frac {f(-3b/2)}{g(-q/2)}=\frac{f(-3b/2)/(-3b/2)}{g(-q/2)/(-q/2)}\cdot\frac {3b} {q} $$ The first fraction tends to $1/2$ and hence the limit in question is equal to the limit of $3b/2q$.
Now $$\frac{b} {q} =\left(\frac{3-\sqrt{7}}{2-\sqrt{2}}\right)^n\to 0$$ and hence desired limit is $0$.