Computing $\lim_{x\to 0^+}\frac{\ln(x)}{\ln(\tan^2(x))}$

Question

I have to find : $$l=\lim\limits_{x \to 0^{+}}\dfrac{\ln(x)}{\ln(\tan^2(x))}$$ As both $\ln(x)$ and $\ln(\tan^2(x))$ $\rightarrow 0$ when $x \rightarrow 0^{+}$ I can use L'Hopital's rule. We have that: $$ \begin{align} \left(\frac{\ln(x)}{\ln(\tan^2(x))}\right)' =& \frac{1/x}{1/(\sin^2(x)/\cos^2(x)) \cdot 2 \cdot (\sin(x)/\cos(x)) \cdot 1/(\cos^2(x))} \\ =& \frac{1/x}{2/(\cos(x)\sin(x))}\\ =&\frac{\cos(x)\sin(x)}{2x} \end{align} $$

and if I differentiate this once again I have that

$$\frac{-\sin^2(x)+\cos^2(x)}{2} \rightarrow \frac{1}{2}$$ when $x \rightarrow 0^{+}$

Is this correct? Thanks!

Answer 1

As pointed out by Luke the given limit is in $\frac {-\infty} {-\infty}$ form.

Otherwise what you have done is correct but you don't have to apply L'Hopital's rule a second time. Just use the facts that $\cos x \to 1$ and $\frac {\sin x } x \to 1$.

Answer 2

$$\lim_{x \rightarrow 0^{+} } \frac{\ln(x)}{\ln(\tan^2(x))} = \lim_{x \rightarrow 0^{+} } \frac{\ln(x)}{2\ln(\tan(x))} = \frac{1}{2} \lim_{x \rightarrow 0^{+} } \frac{\ln(x)}{\ln(\tan(x))} = \text{L'Hospital's rule} = \\ \frac{1}{2} \lim_{x \rightarrow 0^{+} } \frac {\frac {1}{x}}{\frac {1}{\tan x} \frac{1}{\cos^2x}} = \frac{1}{2} \lim_{x \rightarrow 0^{+} } \frac {\sin x \cos x}{x} = \frac{1}{2} \lim_{x \rightarrow 0^{+} } \frac {\sin x}{x} \lim_{x \rightarrow 0^{+} } {\cos x} = \frac{1}{2}.$$ $$$$

An edited answer. Thanks for the comment @LostInSpace

Answer 3

$$L=\lim_{x\to 0^+}\frac{\ln x}{\ln\left(\tan ^2 x\right)}=\frac{1}{2}\lim_{x\to 0^+}\frac{\ln x}{\ln\left(\tan x\right)}=\frac{1}{2}\lim_{x\to 0^+}\left(1+\frac{\ln\left(\frac{x}{\tan x}\right)}{\ln(\tan x)}\right)$$

As already mentioned:

$$\color{red}{\text{standard limit from the table:}}\;\lim_{x\to 0}\frac{x}{\tan x}=1$$ $$\lim_{x\to 0}\ln(\tan x)=-\infty$$ $$\implies\lim_{x\to 0^+}\frac{\ln\left(\frac{x}{\tan x}\right)}{\ln(\tan x)}=0\implies L=\frac{1}{2}$$

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