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  • If [.] denotes the greatest integer function then <img alt="\mathrm{\lim _{n \rightarrow \infty} \frac{[x]+[2 x]+\ldots+[n x]}{n^2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cma

If [.] denotes the greatest integer function then \mathrm{\lim _{n \rightarrow \infty} \frac{[x]+[2 x]+\ldots+[n x]}{n^2}} is 

Option: 1

0


Option: 2

\mathrm{x}


Option: 3

\mathrm{\frac{x}{2}}


Option: 4

\mathrm{\frac{x^{2}}{2}}


Answers (1)

best_answer

\mathrm{n x-1<[n x] \leq n x. Putting n=1,2,3, \ldots n \: and \: adding \: them,}\mathrm{x \Sigma n-n<\Sigma[n x] \leq x \Sigma n}

\mathrm{\therefore \quad x \cdot \frac{\Sigma n}{n^2}-\frac{1}{n}<\frac{\Sigma[n x]}{n^2} \leq x \cdot \frac{\Sigma n}{n^2}}\: \: \: \: \: \: \: \: .....(1)

\mathrm{\text { Now, } \lim _{n \rightarrow \infty}\left\{x \cdot \frac{\Sigma n}{n^2}-\frac{1}{n}\right\}=x \cdot \lim _{n \rightarrow \infty} \frac{\Sigma n}{n^2}-\lim _{n \rightarrow \infty} \frac{1}{n}=\frac{x}{2}}

              \mathrm{\lim _{n \rightarrow \infty}\left\{x \cdot \frac{\Sigma n}{n^2}\right\}=x \cdot \lim _{n \rightarrow \infty} \frac{\Sigma n}{n^2}=\frac{x}{2}}

As the two limits are equal , by (1) \mathrm{\lim _{n \rightarrow \infty} \frac{\Sigma[n x]}{n^2}=\frac{x}{2}}

Note Here we have to select the correct option. In many cases we can adopt the method of elimination by trial. As the answer is to be correct for all x, take x=1. Then

\mathrm{\lim _{n \rightarrow \infty} \frac{[x]+[2 x]+\ldots+[n x]}{n}=\lim _{n \rightarrow \infty} \frac{1+2+\ldots+n}{n^2}=\frac{1}{2}}

When x=1 is put in the options, we find that (a), (b) are wrong.

Now, take x=2. Then

\mathrm{\lim _{n \rightarrow \infty} \frac{[x]+[2 x]+\ldots+[n x]}{n^2}=\lim _{n \rightarrow \infty} \frac{2+4+\ldots+2 n}{n^2}=1}

For x=2, in the options (c), (d), we find that (d) is also wrong. If the option (d) is given as "none of these" then this process of elimination becomes unsafe unless all other options are verified to be incorrect.

Posted by

Pankaj

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