Calculate limit with or without the use of l'Hospital's rule or series expansions $$\lim_{x\to0^{+}}\left(\frac{\cos^2x}{x}-\frac{e^x}{\sin x}\right)$$
One can easily calculate this limit by using series expansions for the functions appearing inside the round brackets, yielding $-1$.
My question is: can anyone give another proof of this result without the use of either l'Hospital's rule or series expansions?

$\begin{array}\\
\dfrac{\cos^2x}{x}-\dfrac{e^x}{\sin x}
&=\dfrac{1-\sin^2x}{x}-\dfrac{e^x}{x+O(x^3)}\\
&=\dfrac{1-(x+O(x^3))^2}{x}-\dfrac{1+x+O(x^2)}{x+O(x^3)}\\
&=\dfrac{1-x^2+O(x^3)}{x}-\dfrac{1+x+O(x^2)}{x}(1+O(x^2))\\
&=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)(1+O(x^2))}{x}\\
&=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)}{x}\\
&=\dfrac{-x-x^2+O(x^3)}{x}\\
&=-1-x+O(x^2)\\
&\to -1\\
\end{array}
$
