Evaluate $\lim_{x\rightarrow\frac{\pi}{2}^\pm}{\log{\left({\frac{1}{\cos^4x}}\right)}-\tan^2{x}}$ I have the following function:
$$f(x)=\log{\left({\frac{1}{\cos^4x}}\right)}-\tan^2{x}$$
The function is defined in $\left({-\frac{\pi}{2},\frac{\pi}{2}}\right)$. I want to study the behaviour of the function when $x\rightarrow \frac{\pi}{2}^\pm$:
$$\lim_{x\rightarrow\frac{\pi}{2}^\pm}{\log{\left({\frac{1}{\cos^4x}}\right)}-\tan^2{x}}$$
We have $\log{\left({\frac{1}{\cos^4x}}\right)}\rightarrow+\infty$ because $\frac{1}{\cos^4{x}}\rightarrow+\infty$ and $\tan^2{x}\rightarrow+\infty$.
Therefore shouldn't this all lead to a form of indertermination $[\infty-\infty]$? My textbook reports that the limit is actually $-\infty$ for both $x\rightarrow\frac{\pi}{2}^+$ and $x\rightarrow\frac{\pi}{2}^{-}$. I'm very confused as to how to calculate these limits. Any hints?
 A: This might be an overkill, but you could write
$$ \tan^2 x = \log (\exp (\tan^2x)),$$
so
\begin{align}
f(x) &= \log\left(\frac{1}{\cos^4 x}\right) - \tan^2x\\
&= \log\left(\frac{1}{\cos^4 x}\right) - \log (\exp (\tan^2x))\\
&= \log\left(\frac{1}{\exp(\tan^2 x)\cos^4 x}\right).
\end{align}
Now do a variable change $u = \cos^2x$, so that $\sin^2x = 1-u$, and $u \to 0^+$ when $ x \to \pm \pi/2.$ This gives
$$ \lim_{x\to\pm \pi/2} f(x) = \lim_{u\to 0^+} \log\left( \frac{1}{\exp\left(\frac{1}{u}-1\right)u^2} \right) \to \log\left(\frac{1}{+\infty}\right) = \log(0^+) = -\infty, $$
since the logarithm is continuous, and $\exp(1/u-1)\to +\infty$ faster than $u^2\to 0^+$ as $u \to 0^+$.
A: Think about how fast each of the two parts of the function tend to infinity, will one dominate the other?
A: Note that
$$f(x)=\log(\sec^4x)-(\sec^2x-1)=1+2\log u-u$$
where $u=\sec^2x\to\infty$ as $x\to\pi/2$. But $\lim_{u\to\infty}(1+2\log u-u)=-\infty$ is (relatively) easy to see, since ${\log u\over u}\to0$ as $u\to\infty$. So
$$\lim_{x\to\pi/2}\left(\log\left(1\over\cos^4x\right)-\tan^2x\right)=-\infty$$
Remark: Since the two-sided limit diverges to $-\infty$, so do each of the one-sided limits.
A: Let $x=\frac{\pi}2-y$ with $y\to 0$ therefore
$$\lim_{x\rightarrow\frac{\pi}{2}^\pm}{\log{\left({\frac{1}{\cos^4x}}\right)}-\tan^2{x}}
=
\lim_{y\to 0}\, {-\log{\left({\sin^4y}\right)}-\frac1{\tan^2{y}}}\to -\infty$$
indeed by  $\sin^2x=t \to 0^+$
$$-\log{\left({\sin^4y}\right)}-\frac{\cos^2 y}{\sin^2{y}}=-\log t^2-\frac{1-t}{t} =\frac1t\left(-2t\log t-1+t\right)\to \infty\cdot (0-1+0)$$
