\(lim _{ n → ∞}\bigg (\frac{n^2}{(n^2+1)(n+1)}+\frac{n^2}{(n^2+4)(n+2)}+\frac{n^2}{(n2+9)(n+3)}.....+ \frac{n^2}{(n^2+n^2)(n+n)}\bigg)\)is equal to
- \(\frac{π}{8}+\frac{1}{4} \;log_e2\)
- \(\frac{π}{4}+\frac{1}{8}\; log_e2\)
- \(\frac{π}{4}-\frac{1}{8}\; log_e2\)
- \(\frac{π}{8}+\frac{1}{8}\; log_e\sqrt2\)
The Correct Option is A
Solution and Explanation
\(lim _{ n → ∞}\bigg (\frac{n^2}{(n^2+1)(n+1)}+\frac{n^2}{(n^2+4)(n+2)}+\frac{n^2}{(n2+9)(n+3)}.....+ \frac{n^2}{(n^2+n^2)(n+n)}\bigg)\)
= \(lim _{n → ∞} ∑^n_{ r=1} \frac{n^2}{(n^2+r^2)(n+r)}\)
= \(lim_{ n → ∞} ∑^n_{ r=1} \frac{1}{\bigg[1+(\frac{r}{n})^2\bigg]\bigg[1+(\frac{r}{n})\bigg]}\)
= \(∫^1_0 \frac{1}{(1+x^2)(1+x)}dx\)
= \(\frac{1}{2} ∫^1_0\bigg [\frac{ 1}{1+x}-\frac{(x-1)}{(1+x^2)}\bigg]dx\)
= \(\frac{1}{2}\bigg[\;In(1+x)-\frac{1}{2}\;In(1+x^2)+tan^{-1}x\bigg]^1_0\)
= \(\frac{1}{2}\bigg[\frac{π}{4}+\frac{1}{2}\;In2\bigg]\)
\(=\frac{π}{8}+\frac{1}{4}\;In2\)
Hence, the correct option is (A): \(\frac{π}{8}+\frac{1}{4} \;log_e2\)
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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).