If $f$ is continuous on $[0,1]$ and $f\big(\large\frac{1}{3}\big)$$=1 then \lim\limits_{n\to \infty}f(\large\frac{n}{\sqrt{9n^2+1}}\big)= (a)\;1\qquad(b)\;0\qquad(c)\;1/3\qquad(d)\;None\;of\;these 1 Answer \lim\limits_{n\to \infty}f\big(\large\frac{n}{\sqrt{9n^2+1}}\big)=$$f\big(\lim\limits_{n\to \infty}\large\frac{n}{\sqrt{9n^2+1}}\big)$
$\Rightarrow f(\large\frac{1}{3})$
$\Rightarrow 1$
Hence (a) is the correct answer.