Question

If $f\left(x\right)$ is continuous in $[0,1]$ and $f\left(\frac{1}{3}\right)=1$ then $\underset{n\to \infty }{lim}f\left(\frac{n}{\sqrt{9n^2+1}}\right)$ is

• A. $1$
• B. $0$
• C. $\frac{1}{3}$
• D. none of these

Answer 1

The correct option is A $1$

$\underset{n\to \infty }{lim}f\left(\frac{n}{\sqrt{9n^2+1}}\right)=\underset{n\to \infty }{lim}f\left(\frac{n}{n\sqrt{9+\frac{1}{n}}}\right)$

$n\to \infty$

$\frac{1}{n}\to 0$

$=f\left(\frac{1}{3}\right)=1$