$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
 A: Simplify before applying the rule...
$$
\lim_{x\to 0} \frac{\tan x- \sin x}{x^3} = \lim_{x\to 0}\frac{\sin x - \sin x \cos x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0}\frac{1-\cos x}{x^2} = \lim_{x\to 0}\frac{1-\cos x}{x^2}
$$
and then apply the rule
$$
 \lim_{x\to 0}\frac{1-\cos x}{x^2}= \lim_{x\to 0}\frac{\sin x}{2x} = \frac 12.
$$
A: It is much easier with Taylor expansion
$$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}=\lim_{x\to 0}\frac{\left(x+\frac{x^3}{3}+\ldots\right)-\left(x-\frac{x^3}{3!}+\ldots\right)}{x^3}=\color{blue}{\frac12}$$
A: You can pull out a $\tan x$ factor and
$$\frac{\tan x-\sin x}{x^3}=\frac{\tan x}x\frac{1-\cos x}{x^2}=\frac{\tan x}x\frac{2\sin^2\dfrac x2}{x^2}.$$
This is enough to conclude
$$\to\frac12.$$

Direct L'Hospital is manageable
$$\frac{\tan x-\sin x}{x^3}\to\frac{\tan^2x+1-\cos x}{3x^2}\to\frac{2\tan x(\tan^2x+1)+\sin x}{6x}\to\frac{2+1}6,$$
but easier after pulling $\tan x$,
$$\frac{\tan x}x\frac{1-\cos x}{x^2}\to\frac{\tan x}x\frac{\sin x}{2x}.$$
A: If you want to go through L'Hospital you may separate $\frac{1}{\cos x}$ the limit of which is $1$ for $x \to 0$:
$$\frac{\tan x - \sin x}{x^3} = \frac 1{\cos x}\cdot \frac{\sin x - \sin x \cos x}{x^3}$$
So, you only need to calculate the limit for $x \to 0$ of
$$\frac{\sin x - \sin x \cos x}{x^3}= \frac{\sin x - \frac 12\sin 2x }{x^3}$$ $$\stackrel{3\times L'Hosp.}{\sim}\frac{-\cos x+4\cos 2x}{6}\stackrel{x \to 0}{\longrightarrow} \frac 12$$
A: God knows why do you want to use L'Hopital rule only. Either you can use brute force Differentiation or you can simply put aside $\cos x$ as others have pointed out, because it is not causing any problem in the limit.
However, a creative way to do this limit-
Let $L= \displaystyle \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$
Now, let $x=3\theta$ , as $ x \rightarrow 0$ , $\theta \rightarrow 0$
$$L= \displaystyle \lim_{\theta \to 0 } \frac{\tan (3 \theta) - \sin(3\theta)}{(3\theta)^3} \\ = \displaystyle \lim_{\theta \to 0 } \dfrac{\frac{3 \tan(\theta)- \tan^3 (\theta)}{1-3\tan^2(\theta)} -(3\sin(\theta)-4\sin^3(\theta))}{27\theta^3} $$
Take the LCM and arrange the terms:
$$ L= \displaystyle \lim_{\theta \to 0} \dfrac{3\tan \theta -3\sin \theta - \tan^3 \theta +4\sin^3 \theta +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3(1-3\tan^2 \theta) }$$
Note that $(1-3\tan^2 \theta)$ is  just $1$ as $\theta \rightarrow 0$, So, we can shift the limit on it so separate it out. (I would appreciate if someone could write that in a better way, I can't)
This limit is now limited to
$$ L=  \displaystyle \lim_{\theta \to 0} \dfrac{3(\tan \theta -\sin \theta) +4\sin^3 \theta - \tan^3 \theta  +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3} $$
Do you feel the Deja Vu?
$$L =  \frac{3L}{27}  +\displaystyle \lim_{\theta \to 0}   \dfrac{4\sin^3 \theta - \tan^3 \theta  +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3}  \\ \implies 24L= 4 \displaystyle \lim_{\theta \to 0} \frac{\sin^3 \theta}{\theta^3} - \displaystyle \lim_{\theta \to 0}  \frac{\tan^3\theta}{\theta^3} + 9 \displaystyle \lim_{\theta \to 0} \frac{\sin \theta \tan^2 \theta }{\theta^3 } -12 \displaystyle \lim_{\theta \to 0}  \frac{\sin^3 \theta \tan^2 \theta}{\theta^3} $$
Hence,  $ L = \frac{ 4 -1 +9}{24} = \frac{1}{2}$
A: I just realized.
$\lim _{x\to 0} \frac{2\cos^{-3} x\sin x+\sin x}{6x}$
$\lim _{x\to 0} \frac{(2\cos^{-3} x + 1)(\sin x)}{6x} = \frac 12$
A: Well you got
$\lim_{x\to 0}\frac{\sec^2x-\cos x}{3x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{\frac{1}{\cos^2x}-\cos x}{x^2}$
$\frac{1}{3}\lim_{x\to 0}\frac{1-\cos^3x}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x+\cos^2x)}{x^2\cos^2x}$
$\frac{1}{3}\lim_{x\to 0}\frac{(1+\cos x+\cos^2x)}{\cos^2x}\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\frac{1}{3}\frac{(1+1+1)}{1}\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\frac{1}{3}3\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
$\lim_{x\to 0}\frac{(1-\cos x)}{x^2}$
Multiplying the numerator and denominator of the limit by $(1+\cos x)$ we get
$\lim_{x\to 0}\frac{(1-\cos x)(1+\cos x)}{x^2}\frac{1}{(1+\cos x)}$
$\lim_{x\to 0}\frac{(1-\cos^2x)}{x^2}\lim_{x\to 0}\frac{1}{(1+\cos x)}$
We know $\sin^2x+\cos^2x=1$
From this we get $(1-\cos^2x)=\sin^2x$
$\lim_{x\to 0}\frac{\sin^2x}{x^2}\frac{1}{2}$
$\frac{1}{2}\lim_{x\to 0}\frac{\sin^2x}{x^2}$
$\frac{1}{2}[\lim_{x\to 0}\frac{\sin x}{x}]^2$
Using L'Hopital's rule to evaluate the limit we get,
$\frac{1}{2}[\lim_{x\to 0} \frac{\cos x}{1}]^2$
$\frac{1}{2}(1)^2$
$\frac{1}{2}$
