LMVT Application: $\sqrt {n + 1} - \sqrt n

LMVT Application: $\sqrt {n + 1} - \sqrt n < \frac{1}{{2N}}$

Problem

${\rm{Prove that, }}\sqrt {n + 1} - \sqrt n {\rm{ < }}\frac{1}{{2N}}{\rm{ if }}n > {N^2},{\rm{ n being a natural number}}{\rm{. }}$

Solution

${\rm{Let }}f(x) = \sqrt x ,x \in [n,n + 1]$

$\begin{array}{l} {\rm{Since n is a natural number, clearly f(x) satisfies the continuity and}}\\ {\rm{ differentiability requirements of LMVT}}{\rm{.}} \end{array}$

$f'(c) = \frac{1}{{2\sqrt c }} = \frac{{f(n + 1) - f(n)}}{{(n + 1) - n}},n < c < n + 1$

$\frac{1}{{2\sqrt c }} = f(n + 1) - f(n) = \sqrt {n + 1} - \sqrt n .........(*)$

${\rm{Since, }}n < c < n + 1$

$\sqrt n < \sqrt c < \sqrt {n + 1}$

$\frac{1}{{\sqrt n }} > \frac{1}{{\sqrt c }} > \frac{1}{{\sqrt {n + 1} }}..........(\# )$

${\rm{Given, }}n > {N^2}$

$\sqrt n > N$

$\frac{1}{{\sqrt n }} < \frac{1}{N},\frac{1}{{\sqrt n }} > \frac{1}{{\sqrt c }}{\rm{ from (\# )}}$

${\rm{Hence, }}\frac{1}{N} > \frac{1}{{\sqrt c }}$

$\frac{1}{{2N}} > \frac{1}{{2\sqrt c }}$

$\frac{1}{{2N}} > \sqrt {n + 1} - \sqrt n {\rm{ from (*)}}$

123IITJEE by IIT Alumnus Manish Verma

Search This Blog

Updates ...

LMVT Application: $\sqrt {n + 1} - \sqrt n < \frac{1}{{2N}}$

Popular posts from this blog