a simple problem on limit made me confused

Question

I tried to solve this problem : $$\lim_{x \to \frac{\pi}{2}} (1+\cos x)^{\tan x}$$

if $y=(1+\cos x)^{\tan x} \implies \log y=\tan x \log(1+\cos x)$

$$\lim_{x \to \frac{\pi}{2}}\log y=\lim_{x \to \frac{\pi}{2}}[\tan x\log(1+\cos x)]=\lim_{x \to \frac{\pi}{2}}\frac{\log(1+\cos x)}{\cos x} \times \lim_{x \to \frac{\pi}{2}} \sin x$$$$=\lim_{x \to \frac{\pi}{2}} \frac{(-\sin x)}{(1+\cos x)(-\sin x)}=\lim_{x \to \frac{\pi}{2}} \frac{1}{1+\cos x}=1$$

$$\implies\log\lim_{x \to \frac{\pi}{2}}(1+\cos x)^{\tan x}=1$$

$$\implies\lim_{x \to \frac{\pi}{2}}(1+\cos x)^{\tan x}=e$$

I got answer $"e"$ but in my book it is $"e^{-1}"$. So I am confused whether I am right , please solve this, it will be definitely appreciable

Answer 1

you can let $y=\lim_{x \to \frac{\pi}{2}} (1+\cos x)^{\tan x}$ then $$\ln y=\ln \lim_{x \to \frac{\pi}{2}} (1+\cos x)^{\tan x} \\ \ln y= \lim_{x \to \frac{\pi}{2}} \ln(1+\cos x)^{\tan x} \\ \ln y= \lim_{x \to \frac{\pi}{2}} \tan x\cdot\ln(1+\cos x) \\ \ln y= \lim_{x \to \frac{\pi}{2}} \frac{\ln(1+\cos x)}{\cot x} \\ \ln y= \lim_{x \to \frac{\pi}{2}} \frac{\frac{-\sin x}{(1+\cos x)}}{-\csc^2 x} \\ \ln y= \lim_{x \to \frac{\pi}{2}} \frac{\sin^3 x}{(1+\cos x)}=1 \\ \implies y=e $$

Answer 2

As $x \to \pi/2, (1 + \cos x)^{\tan x} = [(1 + \cos x)^ {(1/cosx)}]^{(\sin x)} \to e^{(\sin(\pi/2))} = e^1 = e $ !

Answer 3

It is trivial to see that the limit, should it exist, must be at least $1$, since for any $0 < x < \pi/2$, we have $(1+\cos x)^{\tan x} > 1^{\tan x} = 1$. Therefore, the book's answer of $1/e < 1$ cannot possibly be correct.

Answer 4

Built around $x=\frac{\pi }{2}$ , the Taylor series of $(1+\cos x)^{\tan x}$ is $$e+\frac{1}{2} e \left(x-\frac{\pi }{2}\right)-\frac{1}{24} e \left(x-\frac{\pi }{2}\right)^2-\frac{7}{48} e \left(x-\frac{\pi }{2}\right)^3+O\left(\left(x-\frac{\pi }{2}\right)^4\right)$$

This come from the fact that the Taylor series (built at $x=\frac{\pi }{2}$ are, respectively for $(1+\cos x)$ and $\tan x$, $$A=1-\left(x-\frac{\pi }{2}\right)+\frac{1}{6} \left(x-\frac{\pi }{2}\right)^3-\frac{1}{120} \left(x-\frac{\pi }{2}\right)^5+O\left(\left(x-\frac{\pi }{2}\right)^7\right)$$ $$B=-\frac{1}{x-\frac{\pi }{2}}+\frac{1}{3} \left(x-\frac{\pi }{2}\right)+\frac{1}{45} \left(x-\frac{\pi }{2}\right)^3+\frac{2}{945} \left(x-\frac{\pi }{2}\right)^5+O\left(\left(x-\frac{\pi }{2}\right)^7\right)$$ Now, compose the series for $A^B$ (using logarithms make things slightly easier).