<s>[INST] Is there any way to evaluate this limit without applying de l'Hôpital rule nor series expansion? Is there a way to evaluate this limit:
$$\lim_{x \to 0}  \frac{\sin(e^{\tan^2 x} - 1)}{\cos^{\frac35}(x) - \cos(x)}$$
without using de l'Hôpital rule and series expansion?  
Thank you,
 [/INST] This is similar to the accepted answer by user Pacciu, but avoids the use of "asymptotics" and $\approx$ symbols as these are totally unnecessary and perhaps can be confusing to a beginner. Rules of limits combined with the following standard limits is sufficient: $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ where $n$ is rational.
We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\sin(e^{\tan^{2}x} - 1)}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{x \to 0}\frac{\sin(e^{\tan^{2}x} - 1)}{e^{\tan^{2}x} - 1}\cdot\frac{e^{\tan^{2}x} - 1}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{t \to 0}\frac{\sin t}{t}\cdot\lim_{x \to 0}\frac{e^{\tan^{2}x} - 1}{\cos^{3/5}x - \cos x}\text{ (putting }t = e^{\tan^{2}x} - 1)\notag\\
&= 1\cdot\lim_{x \to 0}\frac{e^{\tan^{2}x} - 1}{\tan^{2}x}\cdot\frac{\tan^{2}x}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{x \to 0}\frac{\tan^{2}x}{\cos^{3/5}x - \cos x}\text{ (putting }t = \tan^{2}x)\notag\\
&= 1\cdot\lim_{x \to 0}\frac{\sin^{2}x}{\cos^{2}x}\cdot\frac{1}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{1\cdot\{\cos^{3/5}x - \cos x\}}\notag\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{x^{2}}{\cos^{3/5}x - \cos x}\notag\\
&= 1\cdot\lim_{x \to 0}\frac{x^{2}}{(1 - \cos x) - (1 - \cos^{3/5}x)}\notag\\
&= \lim_{x \to 0}\frac{x^{2}}{(1 - \cos x)}\cdot\dfrac{1}{1 - \dfrac{1 - \cos^{3/5}x}{1 - \cos x}}\notag\\
&= 2\lim_{x \to 0}\dfrac{1}{1 - \dfrac{\cos^{3/5}x - 1}{\cos x - 1}}\notag\\
&= 2\lim_{t \to 1}\dfrac{1}{1 - \dfrac{t^{3/5} - 1}{t - 1}}\text{ (putting }t = \cos x)\notag\\
&= 2\cdot\dfrac{1}{1 - \dfrac{3}{5}} \text{ (here }n = 3/5, a = 1)\notag\\
&= 5\notag
\end{align}
</s>