Compute the following limits if they exist lim_{xrightarrow0}frac{sec x-1}{x^3}

Given:
$\underset{x\to 0}{lim}\frac{\sec x-1}{x^3}$
If the value of $\underset{x\to a}{lim}\frac{f(x)}{g(x)}$ is in indeterminate form like $\frac{0}{0},\frac{\mathrm{\infty }}{\mathrm{\infty }},{\mathrm{\infty }}^0$, etc, then $\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$
Explanation:
Here $f(x)=\sec x-1$ and $g(x)=x^3$
By plugging the value of the limit x= 0,
$f(x)=\sec (0)-1=1-1=0$
$g(x)=(0^3)=0$
So, the indeterminate form is $\frac{0}{0}$
The derivative of f(x) is $f^{\prime }(x)=\sec (x)\tan (x)$
The derivative od the is $g^{\prime }(x)=3x^2$
$\underset{x\to 0}{lim}\frac{\sec x-1}{x^3}=\underset{x\to 0}{lim}\frac{\sec x\cdot \tan x}{3x^2}$
Differentiate f(x) and g(x) with respect to x,
$f^{″}(x)=-\sec (x)+2{\sec }^3(x)$ and $g^{″}(x)=6x$
$\underset{x\to 0}{lim}\frac{\sec x-1}{x^3}=\underset{x\to 0}{lim}\frac{-\sec x+2{\sec }^3(x)}{6x}$
Take $\frac{1}{6}$ outside and rewrite this as
$\frac{1}{6}\underset{x\to 0}{lim}(-\sec (x)+2{\sec }^3(x))x^{-}1$
As x approaches 0 from the left side the above limit goes to negative infinity.
So it is, $\frac{1}{6}-\mathrm{\infty }=-\mathrm{\infty }$
Thus, $\underset{x\to 0}{lim}\frac{\sec x-1}{x^3}=-\mathrm{\infty }$