The value of $\displaystyle\lim_{n \to\infty} \frac{1+2+3+...n}{n^{2}+100}$ is equal to :
- $\infty $
- $\frac{1}{2}$
- 2
- 0
The Correct Option is B
Solution and Explanation
$ =\displaystyle\lim_{n\to\infty} \frac{n\left(n+1\right)}{\left(n^{2}+100\right)} $
(By using sum of $n$ natural number $1+ 2 + 3 + .... + n = \frac{n\left(n+1\right)}{2} $)
Take $n^2$ common from $N^r$ and $D^r$.
$ =\displaystyle\lim_{n\to\infty} \frac{n^{2} \left(1+ \frac{1}{n}\right)}{2n^{2}\left(1+ \frac{100}{n^{2}}\right) } = \frac{1}{2} $
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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).