Question

lim(x-0) {sin2x+sin3x}/ {2x+sin3x} .....evaluate

Answer 1

$all \: steps \: limit \: x \: tends \: to \: zero \\ \frac{}{2x + sin3x} \\ dividing \: by \: x \\ \frac{\frac{sin2x + sin3x}{x} }{ \frac{2x + sin3x}{x} } \\ seperating \\ \frac{ \frac{2sin2x}{2x} + \frac{3sin3x}{3x} }{2 + \frac{3sin3x}{3x} } \\ = \frac{2 + 3}{2 + 3} = 1$

Answer 2

refer to the image given below:-