Question

${lim}_{x\to 0}\frac{sinax+bx}{ax+sinbx}$

Answer 1

${lim}_{x\to 0}\frac{sinax+bx}{ax+sinbx}$

$={lim}_{x\to 0}\frac{\frac{sinax}{x}+b}{a+\frac{sinbx}{x}}=\frac{{lim}_{x\to 0}\frac{sinax}{ax}\times a+b}{{lim}_{x\to 0}\frac{sinbx}{bx}\times b}=\frac{a+b}{a+b}=1$