\(lim _{ n → ∞}\bigg (\frac{n^2}{(n^2+1)(n+1)}+\frac{n^2}{(n^2+4)(n+2)}+\frac{n^2}{(n2+9)(n+3)}.....+ \frac{n^2}{(n^2+n^2)(n+n)}\bigg)\)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): \(\frac{π}{8}+\frac{1}{4} \;log_e2\)
Top Questions on Limits
Consider two statements:
Statement 1: $ \lim_{x \to 0} \frac{\tan^{-1} x + \ln \left( \frac{1+x}{1-x} \right) - 2x}{x^5} = \frac{2}{5} $
Statement 2: $ \lim_{x \to 1} x \left( \frac{2}{1-x} \right) = e^2 \; \text{and can be solved by the method} \lim_{x \to 1} \frac{f(x)}{g(x) - 1} $- Find the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]
- Find the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]
- Find the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]
- Find the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]
Questions Asked in JEE Main exam
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
- JEE Main - 2025
- Sequences and Series
- Power of point source is 450 watts. Radiation pressure on a perfectly reflecting surface at a distance of 2 meters is:
- JEE Main - 2025
- Wave optics
- The mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dilute HCl is: (Given molar mass of Mg = 24 g/mol)
- JEE Main - 2025
- Gas laws
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
Consider the following cell: $ \text{Pt}(s) \, \text{H}_2 (1 \, \text{atm}) | \text{H}^+ (1 \, \text{M}) | \text{Cr}_2\text{O}_7^{2-}, \, \text{Cr}^{3+} | \text{H}^+ (1 \, \text{M}) | \text{Pt}(s) $
Given: $ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
At equilibrium: $ \left[ \text{Cr}^{3+} \right]^2 / \left[ \text{Cr}_2\text{O}_7^{2-} \right] = 10^{-7} $
Objective: $ \text{Determine the pH at the cathode where } E_{\text{cell}} = 0. $- JEE Main - 2025
- Electrochemical Cells and Gibbs Free Energy
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).