Question

${lim}_{x\to 0}\frac{3sin2x+2x}{3x+2tan3x}$

Answer 1

${lim}_{x\to 0}\frac{3sin2x+2x}{3x+2tan3x}$,Dividing numerator and denominator by x we get

$={lim}_{x\to 0}\frac{3\frac{sin2x}{x}+2}{3+2tan\frac{3x}{x}}=\frac{{lim}_{x\to 0}\frac{sin2x}{x}+{lim}_{x\to 0}2}{{lim}_{x\to 0}3+{lim}_{x\to 0}\frac{2tan3x}{x}}=\frac{(3{lim}_{2x\to 0}\frac{sin2x}{2x}\times 2)+2}{3+(2{lim}_{3x\to 0}\frac{tan3x}{3x}\times 3)}$

$=\frac{(3\times 2)+2}{3+(2\times 3)}[\because {lim}_{x\to a}\frac{sinx}{x}=1]$

$=\frac{8}{9}$