I came across this question the other day and have been trying to solve it by using some simple algebraic manipulation without really delving into L'Hospital's Rule or the Power Series as I have just started learning limit calculations. We needed to find : $$\lim_{x \to 0} \frac {x\cos x - \sin x}{x^2\sin x}$$ I approached this problem in two different ways and know what the flaw is, however I have been unable to justify why this is so.
Let $$f(x) = \frac {x\cos x - \sin x}{x^2\sin x}$$ Therefore, dividing by $x$, $$f(x) = \frac {\cos x - \frac{\sin x}{x}}{x\sin x}$$ Using standard limit properties, $$\lim_{x \to 0}f(x) = \frac{\lim_{x \to 0}\cos x - \lim_{x \to 0}\frac{\sin x}{x}}{\lim_{x \to 0}x\sin x}$$ Since $$\lim_{x \to 0} \frac {\sin x}{x}=1$$ $$\lim_{x \to 0}f(x)= \frac{\lim_{x \to 0}\cos x-1}{\lim_{x \to 0}x\sin x}$$
Rewriting the above as $$\lim_{x \to 0}\frac{(\cos x -1)x}{x^2\sin x}$$ and using the fact that $\lim_{x \to 0} \frac {\sin x}{x}=1$ and $\lim_{x \to 0}\frac{\cos x -1}{x^2}= -\frac{1}{2}$, we get $$\lim_{x \to 0}f(x)=-\frac{1}{2}$$
I know that the answer is wrong although I am not able to understand why. I believe it is because I cannot combine the numerator and denominator into a single limit function. Using a similar trick, I also obtained the limit to be $-\frac{3}{8}$.
Questions:
1) Could someone please explain why combining the numerator and denominator into a single limit is wrong? (The reason I even went ahead with such a manipulation was, we are allowed to separate the numerator and denominator while expanding the limit of a rational function so I felt that the reverse should also work).
2) As you can notice, I have not used L'Hospital's Rules or Power Series expansion of $\sin x $and $\cos x$. When I used L'Hospital's Rule, I noticed that I needed to go upto the third or fourth derivative to get rid of the $\frac{0}{0}$ indeterminate form. So would there be a better way of approaching such limits?
Thank You.