The value of \(\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{x\cos x - \log (1 + x)}}{{{x^2}}}} \right]\) is:

Calculation:

Given, \(\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{x\cos x - \log (1 + x)}}{{{x^2}}}} \right]\)

which is in the form \(\left(\dfrac{0}{0}\right)\)

using L-Hospital Rule:

on differentiating N^r and ω^r

\(= lim_{x\to 0}\left[\dfrac{cosx -x sinx - \frac{1}{1+x}}{2x}\right]\)

again differentiating w.r.to x

\(= lim_{x\to 0}\left[\dfrac{-sinx -( sinx + xcos x)+\frac{1}{(1+x)^2}}{2}\right]\)

put x → 0

\(= \dfrac{-sin 0^\circ - ( sin 0^\circ + 0. cos 0^ \circ)+ \frac{1}{1+0}}{2}\)

= 1/2