Question

The value of $\underset{x\to 0}{lim}\frac{x\cos x-\log (1+x)}{x^2}$, is

• A. 1
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

The correct option is C $\frac{1}{2}$

Substituting gives us indeterminate form.
So, applying L'Hospitals Rule,
=${lim}_{x\to 0}\frac{-xsinx+cosx-\frac{1}{1+x}}{2x}$
Substituting gives us indeterminate form so, applying L'Hospitals Rule again,
=${lim}_{x\to 0}\frac{-xcosx-sinx-sinx+\frac{1}{(1+x{)}^2}}{2}$
=$\frac{0-0-0+\frac{1}{1+0}}{2}$
=$\frac{1}{2}$