# Unable to solve this limit

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

What I've tried so far:

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

$=1+\sin^2\dfrac{x}{2}$

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!

• – A---B Mar 11 '17 at 19:43
• Might be helpful : math.stackexchange.com/questions/387333/… – A---B Mar 11 '17 at 19:46
• You want this solved without L'Hospital? – user261263 Mar 12 '17 at 11:20
• @Eugene Covaci yes – Abhishekstudent Mar 12 '17 at 12:54

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

$$\frac{1}{60}$$

• I am not familiar with that. Is there an alternative? – Abhishekstudent Mar 11 '17 at 19:21
• @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! – Von Neumann Mar 11 '17 at 19:22
• Sure! Let me do that. – Abhishekstudent Mar 11 '17 at 19:23