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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.
answered Dec 30, 2013 by sreemathi.v
 

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