Evaluate $\lim _ { x \rightarrow 0} \frac { \tan ( x ^ { a } ) - ( \sin x ) ^ { a } } { x ^ { a + 2} }$ For $a \geq 1$, evaluate $\lim _ { x \rightarrow 0} \frac { \tan ( x ^ { a } ) - ( \sin x ) ^ { a } } { x ^ { a + 2} }$
I would like to solve this without Taylor's expansion. I've tried adding and subtracting $x^a$ but I don't know if this is a correct approach. I've also tried L'Hospital and reached the result $(a^2 - a) /(6a +12). $
 A: if $a=1$,
$$\tan (x)=x+\frac {x^3}{3}(1+\epsilon_1 (x)) $$
$$\sin (x)=x-\frac {x^3}{6}(1+\epsilon_2 (x)) $$
the limit is $\frac {1}{2} $.
if $a>1$
$$(\sin (x))^a=e^{a(\ln (x)+\ln (1-\frac {x^2}{6}(1+\epsilon_2(x)))} $$
$$=x^a(1-\frac {ax^2}{6}(1+\epsilon_2 (x)) )$$
the limit is $$\frac {a}{6} $$
A: If $a$ is a positive integer then there is an easy approach via factorization. Let us first deal with $a=1$. We have $$\frac{\tan x - \sin x}{x^{3}} =\frac{\sin x} {x} \frac{1}{\cos x} \frac{1-\cos x} {x^{2}}\to \frac{1}{2}\tag{1}$$ Also note that using L'Hospital's Rule or Taylor series we can easily show that $$\frac{\tan x- x} {x^{3}}\to \frac{1}{3}\tag{2}$$ and subtracting these above two limits we see that $$\frac{x-\sin x} {x^{3}}\to\frac{1}{6}\tag{3}$$ For general integer $a>1$ we have $$\frac{\tan x^{a} - x^{a}} {x^{a+2}} = \frac{\tan x^{a} - x^{a}} {x^{3a}} \cdot x^{2(a-1)}\to\frac{1}{3}\cdot 0=0\tag{4}$$ and $$\frac{x^{a} -\sin^{a} x} {x^{a+2}}=\frac{x-\sin x} {x^{3}}\sum_{i=0}^{a-1}\frac{x^{a-1-i}}{x^{a-1-i}}\frac{\sin^{i}x}{x^{i}}\to\frac{1}{6}\cdot a=\frac{a} {6}\tag{5}$$ Adding equations $(4)$ and $(5)$ we get the desired answer as $a/6$. If $a$ is not an integer then you need to resort to powerful technique of Taylor series as mentioned in another answer. 
