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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?

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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 $u \to 0^+$.

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    $\begingroup$ To put this answer another way, if $u=\sec(x)^2$, then $\left[\ln\left(\frac{1}{\cos\left(x\right)^4}\right)-\tan\left(x\right)^2\right]=1+\ln\left(\frac{u^2}{e^u}\right)$, so the limit is $\lim_{u\to+\infty}1+\ln\left(\frac{u^2}{e^u}\right)=-\infty$. $\endgroup$ – Jam Sep 5 '18 at 12:30
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Think about how fast each of the two parts of the function tend to infinity, will one dominate the other?

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  • $\begingroup$ $\tan^2{x}$ should go faster to infinity $\endgroup$ – Cesare Sep 5 '18 at 11:37
  • $\begingroup$ @Cesare so if the negative amount is going to infinity faster then the function will tend to minus infinity $\endgroup$ – MRobinson Sep 5 '18 at 11:41
  • $\begingroup$ @MRobinson it makes sense but I see it as a form of indetermination $[\infty-\infty]$. $\endgroup$ – Cesare Sep 5 '18 at 11:54
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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.

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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 $\sin^2x=t \to 0^+$

$$-\log{\left({\sin^4y}\right)}-\frac{\cos^2 y}{\sin^2{y}}=-\log t^2-\frac{1-t}{t} =\frac1t\left(-2t\log t-1+t\right)\to \infty\cdot (0-1+0)$$

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