Help with L'Hospital's Rule for $\lim_{x \to 0}\frac{5x^2}{\ln(\sec x)}$ 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. 
 A: Given that $\lim_{x\to0}\frac{f(x)}{g(x)}, \space f'(x) \space g'(x)$ exists; L'Hospitals Rule says that $$\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}$$
$$\lim_{x\to0}\frac{5x^2}{\ln(\sec x)}$$
$$f'(x) = 10x$$
$$g'(x)=\ln (\sec x)=\frac{1}{\sec x}\cdot \sec x \tan x = \frac{\sec x\tan x}{\sec x}=\tan x$$
$$\therefore \lim_{x\to0}\frac{5x^2}{\ln(\sec x)}=\lim_{x\to0}\frac{10x}{\tan x}$$
Using L'Hospital's Rule again..
$$\lim_{x\to0}\frac{10x}{\tan x}=\lim_{x\to0}\frac{10}{\sec ^2x}=\frac{10}{1}=10$$
Since
$$\sec^2x=\frac{1}{\cos^2 x}=\frac{1}{(\cos 0)^2}=\frac{1}{1}=1$$
A: You clearly don't understand L'Hopsital's Rule clearly. 
If $\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x)=0$ and $\lim_{x\to 0}\frac{f(x)}{g(x)}$ exists, then
$$
\lim_{x\to 0}\frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f'(x)}{g'(x)} 
$$
Now your $f(x)=5x^2$ and $g(x)=\ln(\sec(x))$. And now you can proceed from here. Defining $y=1/\ln(\sec(x))$ does not lead you to the right direction. 
A: $$\begin{align}
\dfrac{5x^2}{\ln (\sec x)} &\to \dfrac{10x}{\frac{\tan x \sec x}{\sec x}} \to\dfrac{10}{\sec^2 x}\\
 \longrightarrow \lim_{x \to 0}\dfrac{5x^2}{\ln (\sec x)} &\equiv \lim_{x \to 0} \dfrac{10}{\sec^2 x}=\frac{10}{1}=10\\
\end{align}$$
A: If $y = \log(\sec x)$, then by the chain rule:
$$y' = \frac{1}{1/\cos x}\left(\frac{1}{\cos x}\right)' = \tan x$$
A: Without L'Hospital's rule, but with Maclaurin series (if you are allowed to use them): expand $\log x \sim x-1$ when $x \to 1$ (in you case it is $\sec x \to 1$) and $\cos x \sim 1-\frac{x^2}{2}+\frac{x^4}{24}$ as $x \to 0$, so you get
$$\lim_{x \to 0} \frac{5x^2}{\frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24}}-1}=\lim_{x \to 0}\frac{5(24-12 x^2+x^4)}{12-x^2}=10
$$
