How to calculate limit without L'Hopital How can I calculate limit without using L'Hopital rule?

$\displaystyle\lim_{x\to 0 }\frac{e^{\arctan(x)}-e^{\arcsin(x)}}{1-\cos^3(x)}$.
Let $\arctan x = t$ so that $x = \tan t$ so that $\sin t = x/\sqrt{1 + x^{2}}$ or $\arctan x = t = \arcsin (x/\sqrt{1 + x^{2}})$.

Now we have

$\displaystyle\begin{aligned}\arctan x - \arcsin x &amp;= \arcsin (x/\sqrt{1 + x^{2}}) - \arcsin x\\
&amp;= \arcsin\left(x\sqrt{\frac{1 - x^{2}}{1 + x^{2}}} - \frac{x}{\sqrt{1 + x^{2}}}\right)\\
&amp;= \arcsin\left\{\frac{x}{\sqrt{1 + x^{2}}}\left(\sqrt{1 - x^{2}} - 1\right)\right\}\\
&amp;= \arcsin\left\{\frac{-x^{3}}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\right\}\\
&amp;= \arcsin y = f(x) \text{ (say)}\end{aligned}$

Next we have

$\displaystyle\begin{aligned}\lim_{x \to 0}\frac{e^{\arctan x} - e^{\arcsin x}}{1 - \cos^{3}x} &amp;= \lim_{x \to 0}e^{\arcsin x}\cdot\frac{e^{f(x)} - 1}{1 - \cos^{3} x}\\
&amp;= \lim_{x \to 0}1\cdot\frac{e^{f(x)} - 1}{f(x)}\cdot\frac{f(x)}{1 - \cos^{3}x}\\
&amp;= \lim_{x \to 0}1\cdot 1\cdot\frac{f(x)}{(1 - \cos x)(1 + \cos x + \cos^{2}x)}\\
&amp;= \frac{1}{3}\lim_{x \to 0}\frac{f(x)}{1 - \cos x}\\
&amp;= \frac{1}{3}\lim_{x \to 0}\frac{\arcsin y}{y}\cdot \frac{y}{1 - \cos x}\\
&amp;= \frac{1}{3}\lim_{x \to 0}\frac{y}{2\sin^{2}(x/2)}\\
&amp;= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot \frac{(x/2)^{2}}{\sin^{2}(x/2)}\\
&amp;= \frac{1}{6}\lim_{x \to 0}\frac{y}{(x/2)^{2}}\cdot 1\\
&amp;= \frac{2}{3}\lim_{x \to 0}\frac{y}{x^{2}}\\
&amp;= \frac{2}{3}\lim_{x \to 0}\frac{-x}{\sqrt{1 + x^{2}}\left(\sqrt{1 - x^{2}} + 1\right)}\\
&amp;= \frac{2}{3}\cdot 0 = 0\end{aligned}$

No Taylor or L'Hospital is required. We just need algebraic and trigonometric manipulation combined with the use of standard limits.
arctan x ~ x-1/3(x^3)+...
arcsin x ~ x+1/6(x^3)+...
e^x~1+x+... 
e^ arctan x -e^ arcsin x ~1+ x-1/3(x^3) -(1+x+1/6(x^3) )~ -3 x^3 /6
1-cos^3 x ~ (1-cos x) (1+cos x +cos^2x )~ x^2 /2 (1+cos x+cos^2x)
put them in limit
you will have   lim =0