Question:
$ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ \sqrt{ n^2 + r^2}} $ equals
$ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ \sqrt{ n^2 + r^2}} $ equals
Updated On: Jun 14, 2022
- 1 + $ \sqrt 5 $
- 1 - $ \sqrt 5 $
- - 1 + $ \sqrt 2 $
- 1 + $ \sqrt 2 $
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The Correct Option is B
Solution and Explanation
Let I = $ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ \sqrt{ n^2 + r^2}} = lim_{n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ n \sqrt{ 1 + (r / n)^2}} $
= $ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{ r / n}{ \sqrt {1 + ( r / n)^2}} $
= $ \int \limits_0^2 \frac{x}{\sqrt{ 1 + x^2}} dx = [ \sqrt{ 1 + x^2 }]_0^2 = \sqrt 5 - 1 $
= $ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{ r / n}{ \sqrt {1 + ( r / n)^2}} $
= $ \int \limits_0^2 \frac{x}{\sqrt{ 1 + x^2}} dx = [ \sqrt{ 1 + x^2 }]_0^2 = \sqrt 5 - 1 $
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).