Limit

Calculus Level pending

$\lim_{n\to \infty} \left(\sqrt{\left\lfloor {\left( 9n-\frac{1}{6} \right)^2}\right\rfloor} - (9n-1)\right) = \frac{a}{b}$

Suppose $a$ and $b$ are positive integers and $\gcd(a,b) =1$ . Find the value of $a+b$ .

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 11.

1 solution

Aditya Narayan Sharma
Apr 16, 2017

Since $[x]\le x$ the term is bounded by $\displaystyle \lfloor{\left(9n-\dfrac{1}{6}\right)^2 \rfloor}\le \left(9n-\dfrac{1}{6}\right)^2$ .

$\displaystyle\begin{aligned} L &=\lim_{n\to \infty} \left(\sqrt{\left\lfloor {\left( 9n-\frac{1}{6} \right)^2}\right\rfloor} - (9n-1)\right) \\ &\le \lim_{n\to\infty} \left(9n-\dfrac{1}{6}-9n+1\right)\\ &=\boxed{\dfrac{5}{6}}\end{aligned}$

You have shown that $L \leq \frac56$ only, how do you know that $L$ must be equal to $\frac56$ ?

Pi Han Goh - 4 years, 1 month ago

Limit

The answer is 11.

1 solution

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