$$ \underset{x\to 0}{\lim} \left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)$$

What I've tried so far:

$2+\cos x=1+(1+\cos x)$


But, I do not think this step is fruitful as I am getting stuck thereafter. Kindly provide some sort of help or hint. Thanks in advance!


Try with Taylor series.

$$\sin(x) \approx x - \frac{x^3}{6}$$

$$\cos(x) \approx 1 - \frac{x^2}{2} + \frac{x^4}{24}$$

The result of the limit will be


  • $\begingroup$ I am not familiar with that. Is there an alternative? $\endgroup$ – Abhishekstudent Mar 11 '17 at 19:21
  • $\begingroup$ @Abhishekstudent Then start with unifying the fractions, and use some trick like a multiplication and a division by $x$ and so on... I'll think about another way in the meanwhile! $\endgroup$ – Von Neumann Mar 11 '17 at 19:22
  • $\begingroup$ Sure! Let me do that. $\endgroup$ – Abhishekstudent Mar 11 '17 at 19:23

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