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?
Best Answer
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 $u \to 0^+$.