Question

Evaluate limit x tends to zero sin 2 x + 3 x upon 2 x + sin 3x

Answer 1

limx→0sinxx=1 we have

sin3(2x)sin2(3x)=(sin(2x)2x)3(sin(3x)3x)2(2x)3(3x)2

then

limx→0sin3(2x)sin2(3x)=1189limx→0x=0

Answer 2

Answer:

$\bf\frac{8}{27}$

Step-by-step explanation:

To find :

$\lim_{x \to 0} \frac{\sin 2x+3x}{2x+\sin 3x}$

Now, At x = 0 both the numerator and denominator gives 0 so it will be in-determinant form

So, Applying L'Hopital rule :

$\lim_{x \to 0} \frac{\sin 2x+3x}{2x+\sin 3x}\\\\\implies \lim_{x \to 0} \frac{2\cos 2x+3}{2+3\cos 3x}\\\\\implies \lim_{x \to 0} \frac{-4\sin 2x}{-9\sin 3x}\\\\\implies \lim_{x \to 0}\frac{-8\cos 2x}{-27\cos 3x}\\\\\implies \text{Now, Taking limit }\\\\\implies \frac{-8\times 1}{-27\times 1}\\\\\implies \frac{8}{27}$