1u cannot apply partial limits, as u have done in step III of ur other method.!
\lim_{x\rightarrow 0} \frac{2Sin3x-Sin6x}{x^3}
By LHospital i am getting -27
But by another method i am getting 0
the other method is
\lim_{x\rightarrow 0} \frac{2Sin3x}{x^3}-\frac{Sin6x}{x^3}
\lim_{x\rightarrow 0} \frac{2*3Sin3x}{3x^3}-\frac{6Sin6x}{6x^3}
Since lim x to 0 Sinx/x = 1
\lim_{x\rightarrow 0} \frac{2*3}{x^2}-\frac{6}{x^2}
Therefore
\lim_{x\rightarrow 0} 0
Therefore the answer is 0.
Pls find my mistake
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