Use l$\text{'}$H$$pital$\text{'}$s Rule to evaluate $\underset{x0}{lim}\frac{1-cosx}{9x^2}.$ Then determine the limit using limit laws and commonly known limits$.$

Use l$\text{'}$H$$pital$\text{'}$s Rule to rewrite the given limit so that it is not an indeterminate form.

$\underset{x0}{lim}\frac{1-cosx}{9x^2}=\underset{x0}{lim}$

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Choose the limit equivalent to the given limit that can be evaluated using limit laws and commonly known limits$.$

A$.\underset{x0}{lim}\frac{1-cosx}{9x^2}=\underset{x0}{lim}[\frac{9+9cosx}{1}]$

B$.\underset{x0}{lim}\frac{1-cosx}{9x^2}=\underset{x0}{lim}[(\frac{sinx}{x})\frac{1}{9+9cosx}]$

c$.\underset{x0}{lim}\frac{1-cosx}{9x^2}=\underset{x0}{lim}[(\frac{sinx}{x})(\frac{sinx}{x})\frac{1}{9+9cosx}]$

D$.\underset{x0}{lim}\frac{1-cosx}{9x^2}=\underset{x0}{lim}[\frac{1si{n}^{2}x}{9x^2}]$

The limit found using either method is $\underset{x0}{lim}\frac{1-cosx}{9x^2}=.$

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$($Simplify your answer.$)$