Question

If $\underset{x\to 0}{\mathrm{Lt}}\frac{x(1+a\cos x)+b\sin x}{x}=1$ , then

• A. $a=b$
• B. $a+b=0$
• C. $2a=b$
• D. None of these

Answer 1

The correct option is B $a+b=0$

${lim}_{x\to 0}\frac{x(1+acosx)+bsinx}{x}=1$

$\frac{0}{0}$ form. So by D L' Hospital.

${lim}_{x\to 0}\frac{(1+acosx)+\left(axsinx\right)+bcosx}{1}=1$

$1+a+b=1$

$a+b=0$