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,
 A: You should know the following limits:
$$\begin{split}\lim_{y\to 0} \frac{\sin y}{y} &= 1\\ \lim_{y\to 0} \frac{e^y-1}{y} &= 1\\ \lim_{y\to 0} \frac{\tan y}{ y} &= 1\\ \lim_{y\to 0} \frac{(1+y)^\theta -1}{y} &= \theta \qquad \text{(}\forall \theta \in \mathbb{R}\text{)}\\
\lim_{y\to 0} \frac{1-\cos y}{y^2} &= \frac{1}{2}\end{split}$$
which can be proved using only elementary Calculus tools (i.e. without any Differential Calculus technique).
These five limits are usually written as asymptotic relations in the following manner:
$$\tag{1} \sin y \approx y$$
$$\tag{2} e^y-1 \approx y$$
$$\tag{3} \tan y \approx y$$
$$\tag{4} (1+y)^\theta -1 \approx \theta\ y$$
$$\tag{5} 1-\cos y \approx \frac{1}{2}\ y^2$$
as $y\to 0$. Using asymptotics (1) - (5) you find:
$$\begin{split} \sin(e^{\tan^ 2 x} - 1) &\approx e^{\tan^2 x}-1 &\quad \text{by (1)}\\
&\approx \tan^2 x &\quad \text{by (2)}\\
&\approx x^2  &\quad \text{by (3)}\end{split}$$
$$\begin{split}\cos^{3/5}(x) - \cos(x) &= \Big(\big(1+(\cos x-1)\big)^{3/5} -1\Big) + \Big(1-\cos x\Big)\\ &\approx \frac{3}{5}\ (\cos x-1) + (1-\cos x) &\text{by (4) with } \theta =3/5\\
&= \frac{2}{5}\ (1-\cos x)\\
&\approx \frac{1}{5}\ x^2 &\text{by (5)}\end{split}$$
hence:
$$\lim_{x \to 0} \frac{\sin(e^{\tan ^ 2 {x}} - 1)}{\cos^{\frac{3}{5}}(x) - \cos(x)} = \lim_{x\to 0} \frac{x^2}{\frac{1}{5}\ x^2}=5\; .$$
A: 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}
A: You can write
$$\sin(e^{\tan^2 x}-1)={\sin(e^{\tan^2 x}-1)\over e^{\tan^2 x}-1}\cdot {e^{\tan^2 x}-1 \over \tan^2 x}\cdot {\tan^2 x\over x^2} x^2=:  g(x)\ x^2$$
with $\lim_{x\to 0} g(x)=1$, but I don't see a similar expansion of the denominator without using somehow that $(1+u)^{3/5}\sim 1+{3\over 5}u$ for $u\to 0$.
