What is the limit of:
$$\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$$
So what I tried is:
$$\lim_\limits{x\to0}{{\sin{x}\over\cos{x}}-\sin{x}\over x^3}=\lim_\limits{x\to0}{{\sin{x}}({1\over\cos x}-1)\over x\cdot x^2}$$
From here, using the rule $\lim_\limits{x\to0}{\sin{x}\over{x}}=1$ it remains to evaluate
$$\lim_\limits{x\to0}{{1\over\cos{x}}-1\over x^2}.$$
I tried changing it a lot of ways but it always gets messier so I'm not sure what to apply here to complete the question.
Best Answer
From your last line, $${{1\over\cos{x}}-1\over x^2}={1-\cos^2(x)\over \cos(x)(1+\cos(x))x^2}={1\over \cos(x)(1+\cos(x))}\cdot \left(\frac{\sin(x)}{x}\right)^2.$$ Can you take it from here?