# 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.

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 $$\sec^2x=\frac{1}{\cos^2 x}=\frac{1}{(\cos 0)^2}=\frac{1}{1}=1$$

• No problem at all. – Zhoe Jan 13 '14 at 21:53

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.

• I see, thanks. I was thinking of it as one function. +1 – Emi Matro Jan 13 '14 at 21:45
• Question: what do you mean by "exists"? How can it exist if it's $0/0$? you mean if they exists separately, $f(x)$ and $g(x)$? – Emi Matro Jan 13 '14 at 21:48
• @user436158 No, for example, $\lim_{x\to 0} 5x^2/2x^2$ exists as a finite number, even though both numerator and denominator approaches $0$ as $x\to 0$. But $\lim_{x\to 0} 5x/2x^2$ goes to infinity, thus the limit does not exist. – MoonKnight Jan 13 '14 at 21:59

If $y = \log(\sec x)$, then by the chain rule: $$y' = \frac{1}{1/\cos x}\left(\frac{1}{\cos x}\right)' = \tan x$$

\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}

• thanks, are you sure the answer is $10$? my notes say $5$ is the answer but maybe they're wrong. Thanks for the input. – Emi Matro Jan 13 '14 at 21:40
• Well, it's how it's turning out to be... – K. Rmth Jan 13 '14 at 21:42
• @user436158 Rmth is right. – MoonKnight Jan 13 '14 at 21:43
• this is clear, thanks +1 – Emi Matro Jan 13 '14 at 21:45

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 $$\lim_{x \to 0} \frac{5x^2}{\frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24}}-1}=\lim_{x \to 0}\frac{5(24-12 x^2+x^4)}{12-x^2}=10$$