Use L'Hospital's Rule to evaluate $\lim_{x \to 0}\dfrac{5x^2}{\ln(\sec x)}$
I know that L'hospital's rule is about differentiating over and over again until you no longer have an indeterminate form.
My try:
$\dfrac{5x^2}{\ln(\sec x)}=\dfrac{1}{\ln(\sec x)}5x^2$
$y=\dfrac{1}{\ln(\sec x)},\;\;\;$take natural logarithm of both sides before differentiating:
$\ln y=\ln \left ( \dfrac{1}{\ln(\sec x)}\right ),\;\;\; \implies \ln y=\ln 1-\ln(\ln(\sec x)),\;\;\;$ now differentiate:
$\dfrac{y^{\prime}}{y}= 0 - ...$
I don't know how to proceed. Can you show please? I know that $\ln(\ln(\sec x))$ will use chain rule, but I don't know how to work it... Thanks.