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

Compute the following limits if they exist $\underset{x\to 0}{lim}\frac{\mathrm{sec}x-1}{{x}^{3}}$
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Delorenzoz
Given:
$\underset{x\to 0}{lim}\frac{\mathrm{sec}x-1}{{x}^{3}}$
If the value of $\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}$ is in indeterminate form like $\frac{0}{0},\frac{\mathrm{\infty }}{\mathrm{\infty }},{\mathrm{\infty }}^{0}$, etc, then $\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$
Explanation:
Here $f\left(x\right)=\mathrm{sec}x-1$ and $g\left(x\right)={x}^{3}$
By plugging the value of the limit x= 0,
$f\left(x\right)=\mathrm{sec}\left(0\right)-1=1-1=0$
$g\left(x\right)=\left({0}^{3}\right)=0$
So, the indeterminate form is $\frac{0}{0}$
The derivative of f(x) is ${f}^{\prime }\left(x\right)=\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)$
The derivative od the is ${g}^{\prime }\left(x\right)=3{x}^{2}$
$\underset{x\to 0}{lim}\frac{\mathrm{sec}x-1}{{x}^{3}}=\underset{x\to 0}{lim}\frac{\mathrm{sec}x\cdot \mathrm{tan}x}{3{x}^{2}}$
Differentiate f(x) and g(x) with respect to x,
${f}^{″}\left(x\right)=-\mathrm{sec}\left(x\right)+2{\mathrm{sec}}^{3}\left(x\right)$ and ${g}^{″}\left(x\right)=6x$
$\underset{x\to 0}{lim}\frac{\mathrm{sec}x-1}{{x}^{3}}=\underset{x\to 0}{lim}\frac{-\mathrm{sec}x+2{\mathrm{sec}}^{3}\left(x\right)}{6x}$
Take $\frac{1}{6}$ outside and rewrite this as
$\frac{1}{6}\underset{x\to 0}{lim}\left(-\mathrm{sec}\left(x\right)+2{\mathrm{sec}}^{3}\left(x\right)\right){x}^{-}1$
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, $\underset{x\to 0}{lim}\frac{\mathrm{sec}x-1}{{x}^{3}}=-\mathrm{\infty }$