When Should I Use Taylor Series for Limits? I get confused between when to apply L'Hospital Rule and Taylor Series.
Is there any set of trigger points in the questions, that would be easier to solve with Taylor Series?
For Example, If the denominator is in terms of a large power of $x (>3)$, then L'Hospital Rule usually becomes complicated and is not advised.
Edit:
Solve 
$\lim _{x\to 0}\left(\left(\sin x\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)$
The options given are: $0, 1, -1, \infty$ 
 A: A general rule to follow is that we always should try at first to apply the simplest method for any limit. Therefore, when we face with an indeterminate form, I suggest at first to try with:
1. Algebraic manipulation, aimed to eliminate the source of that indetermination, e.g.
$$\lim_{x\to 1} \frac{x^2-3x+2}{x^2-1}=\lim_{x\to 1} \frac{(x-1)(x-2)}{(x-1)(x+1)}=\lim_{x\to 1} \frac{x-2}{x+1}=-\frac12$$
2. Standard limits, e.g.
$$x\to 0, \quad \frac{\sin x}x\to 1, \quad \frac{e^x-1}{x}\to 1, \quad x\log x \to 0,\quad \ldots$$
or related results derived from them, see for example: Are all limits solvable without L'Hôpital Rule or Series Expansion.
When we can't solve a limit in an indeterminate form by that basic tools, then we are allowed to use more advanced methods which involve derivatitive concepts and notably l'Hopital rule and Taylor's expansion.
In my opinion, with some special exception and when we are not explicitly requested/forced to use l'Hopital, we always should prefer Taylor's expansion not only because it is a more powerful and effective method but also because it allows us to really understand what is going on and which are the terms which really count in the expression whereas l'Hopital is a blind/black box method which doesn't add any contribution to our knowledge about limits.
Therefore if you know Taylor's expansion, in general, I suggest to proceed always by that.

Edit
The given example is one of the first category that is "solvable by elementary methods". 
Notably we have that as $x \to 0^+$ (otherwise the expression in not defined)
$$\left(\sin x\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}=\left(\sin x\right)^{\frac{1}{x}}+\frac{1}{x^{\sin x}}\to 0 +1=1$$
Indeed we have that


*

*$\left(\sin x\right)^{\frac{1}{x}}\to 0$ (it is in a not-indeterminate form $0^{\infty}$)

*$x^{\sin x}=e^{\sin x \log x}\to e^0=1$
since by standard limits
$$\sin x \log x= \frac{\sin x}x\cdot  x\log x\to 1 \cdot 0 =0$$
A: I should say that most of the time (not to say all the times) Taylor expansions are the keys.
Let me consider your example
$$\lim _{x\to 0}\left(\left(\sin(x)\right)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin(x)}\right)$$
Consider the first term
$$a=\left(\sin(x)\right)^{\frac{1}{x}} \implies \log(a)={\frac{1}{x}}\log\left(\sin(x)\right)$$ Now, by Taylor
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$
$$\log\left(\sin(x)\right)=\log (x)-\frac{x^2}{6}+O\left(x^4\right)$$
$$\log(a)=\frac{\log (x)}{x}-\frac{x}{6}+O\left(x^3\right)$$ So $\log(a)\to -\infty$ which means thet $a\to 0$.
Now, the second term (using equivalent is faster here)
$$b=\left(\frac{1}{x}\right)^{\sin(x)}\implies \log(b)=-\sin(x) \log(x)\sim -x \log(x)\to 0^+$$ which means that $b\to 1^+$ and so $a+b$.
A: When you see, as $f\to0$, the following functions, it is often a good way to use Taylor series rather than L'Hospital rule.


*

*$\log(1+f) = f+o(f)$

*$\sin f = f+o(f)$

*$\cos f = 1+ o(1)$

*$e^f = 1+o(1)$

*$\tan f = f+o(f)$

*$\arctan f = f+o(f)$
(More terms of the series are sometimes necessary). 
