Definite Integration
Integral Squeeze
Grade 12
Question:
<p>If \(I_n=\int_n^{n+1}\dfrac{x}{1+x}\,dx\), find \(\lim_{n\to\infty}I_n\). [JEE Advanced 2012]</p>
<li>\(1\)</li>
<li>\(0\)</li>
<li>\(\dfrac{1}{2}\)</li>
<li>\(\infty\)</li>
Step-by-Step Solution
Key Concept: x/(1+x) = 1 - 1/(1+x). On [n,n+1]: 1/(1+x) \to 0. So Iₙ \to \intₙ^(n+1) 1 dx = 1.
<div class='solution'>
<p>$I_n=\int_n^{n+1}\frac{x}{1+x}dx=\int_n^{n+1}\left(1-\frac{1}{1+x}\right)dx=1-[\ln(1+x)]_n^{n+1}=1-\ln\frac{n+2}{n+1}$</p>
<p>As $n\to\infty$: $\ln\frac{n+2}{n+1}=\ln(1+\frac{1}{n+1})\to 0$. So $I_n\to\boxed{1}$.</p>
</div>
Correct Answer: A