$\displaystyle \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right]$ is equal to
$\log _e2$
- $\log _e\left(\frac{2}{3}\right)$
- 0
- $\log _e\left(\frac{3}{2}\right)$
The Correct Option is A
Solution and Explanation
\(=\lim_{n\to\infty} ∑^n_{r=1}\frac{1}{n}(\frac{1}{1+\frac{r}{n}})\)
\(= ∫_0^1 \frac{1}{1+x}dx=[ℓln(1+x]_0^1 =ℓn2\)
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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).