finding the limit of = $\ln(\frac{2x}{\pi})\cdot e^{\frac{1}{\cos x}}$

Question

I need to find the following limit :

$$ \lim_{x \to \pi/2^-} \ln\left(\frac{2x}{\pi}\right)\cdot e^{\frac{1}{\cos x}}$$

The limit is definitely $-\infty$, yet I don't understand why what I've done is wrong.

$$\cos \left(\frac{\pi}{2} + x \right) =_0 1-\frac{(\pi+2x)^2}{4} +o(x^2)$$ So : $$ \frac{1}{\cos \left(\frac{\pi}{2} + x\right)} =_0 1-\left(1+\frac{(\pi+2x)^2}{4} +o\left(x^2\right)\right)+o(x) = -\frac{(\pi+2x)^2}{4}+o(x)$$ Hence : $$ e^{\frac{1}{\cos (\frac{\pi}{2} + x)}} =_0 1-\frac{(\pi+2x)^2}{4}+o(x)$$

An here what I've obtained is already false because :

$$\lim_{x \rightarrow 0} 1+\frac{(\pi+2x)^2}{4}+o(x) \ne -\infty$$

Answer 1

Let $x=\frac{\pi}2-y$ with $y\to 0^+$ then

$$\displaystyle \lim_{x \rightarrow \frac{\pi}{2}^-} \ln\left(\frac{2x}{\pi}\right)\cdot e^{\frac{1}{\cos x}}=\lim_{y \rightarrow 0^+} \ln\left(1-\frac{\pi y}{2}\right)\cdot e^{\frac{1}{\sin y}}=-\infty$$

indeed

$$\ln\left(1-\frac{\pi y}{2}\right)\cdot e^{\frac{1}{\sin y}}=\frac{\ln\left(1- \frac{\pi y}{2} \right)}{\frac{\pi y}{2}}\cdot \frac{\pi y}{2}e^{\frac{1}{\sin y}}$$

and

$$\frac{\ln\left(1- \frac{\pi y}{2} \right)}{\frac{\pi y}{2}}\to -1$$

$$\frac{\pi y}{2} e^{\frac{1}{\sin y}}=\frac{\pi}{2}\frac{e^{\frac{1}{\sin y}}}{\frac{1}{\sin y}}\frac{y}{\sin y}\to +\infty$$

Answer 2

If you want the solution using Taylor series and equivalents, let, as gimusi did, $x=\frac{\pi}2-y$ with $y\to 0^+$ $$\displaystyle \lim_{x \rightarrow \frac{\pi}{2}^-} \ln\left(\frac{2x}{\pi}\right)\, e^{\frac{1}{\cos (x)}}=\lim_{y \rightarrow 0^+} \ln\left(1-\frac{\pi }{2}y\right)\, e^{{\csc (y)}}$$

Now, using Taylor series built at $y=0$ $$\csc(y)=\frac{1}{y}+\frac{y}{6}+O\left(y^3\right)\implies e^{{\csc (y)}}\sim e^{\frac{1}{y}}$$ $$\ln\left(1-\frac{\pi }{2}y\right)=-\frac{2 y}{\pi }-\frac{2 y^2}{\pi ^2}+O\left(y^3\right)$$ $$\lim_{y \rightarrow 0^+} \ln\left(1-\frac{\pi }{2}y\right)\, e^{{\csc (y)}}\sim -\frac{2}{\pi }\lim_{y \rightarrow 0^+}\left(y\,e^{\frac{1}{y}}\right)=-\infty$$