If , then equals denotes greatest integer function)

JEE MainMathematicsToughLimits, continuity and derivability

If
$\lim_{x \rightarrow 1}\mspace{2mu}\frac{1 - x + lnx}{1 + cos\pi x} = L$ ,
then $|\lbrack 17L\rbrack|$ equals denotes greatest integer
function)

Show answer and solution

\lim_{x \rightarrow 1}\mspace{2mu}\frac{1 - x + lnx}{1 + cos\pi x} = L

Let

x = 1 + h;h \rightarrow 0

L = = \lim_{h \rightarrow 0}\mspace{2mu}\frac{1 - (1 + h) + ln(1 + h)}{1 - cos\pi h} \cdot \frac{\pi^{2}h^{2}}{\pi^{2}h^{2}}

L = \frac{2}{\pi^{2}} \cdot \lim_{h \rightarrow 0}\mspace{2mu}\frac{- h + ln(1 + h)}{h^{2}}\left( \ as\ \lim_{h \rightarrow 0}\mspace{2mu}\frac{\pi^{2}h^{2}}{1 - cos\pi h} = 2 \right)

Put in (1 + h) = t

L = \frac{2}{\pi^{2}} \cdot \lim_{h \rightarrow 0}\mspace{2mu}\frac{t - \left( e^{t} - 1 \right)}{\left( \frac{e^{t} - 1}{t} \right)t^{2}} = - \frac{2}{\pi^{2}} \cdot \lim_{h \rightarrow 0}\mspace{2mu}\frac{e^{t} - 1 - t}{t^{2}} = - \frac{2}{\pi^{2}} \cdot \frac{1}{2} = - \frac{1}{\pi^{2}}

\therefore \mid \lbrack 17L\rbrack = 2

Source: Grass-G / M-09 Mathematics Paper 26 October.docx

Scan any doubt from your notes and get a whiteboard explanation.