Calculate $\lim_{x\to\frac{\pi}{2}^-} (1-\cos x)^{\tan x}$

Question

I tried putting $\cos x = -\frac{1}{\sqrt{1+\tan^2 x}}$, but I am not able to calculate the limit.

Answer 1

Use substitution $t=\frac{\pi}{2}-x$ instead, it leads to $$\lim_{t\to 0^+}(1-\sin t)^{\cot t}$$ Now $\cot t=\frac{\cos t}{\sin t}$ and you can use the fact that $(1+f(t))^{\frac{1}{f(t)}}$ tends to $e$ as $f(t)$ tends to 0.

Answer 2

Hint

1) Set $x=\dfrac\pi2-t$ $\;(t\to 0^-)$.

2) $\ln(1-u)\sim_0 -u$.

Answer 3

Limits like these are often better dealt with by taking logarithms (and then exponentiate the found limit): $$ \lim_{x\to\frac{\pi}{2}^-}\log\bigl((1-\cos x)^{\tan x}\bigr)= \lim_{t\to0^+}-\cot t\log(1+\sin t) $$ where the substitution $t=x-\frac{\pi}{2}$ has been performed.

Now the limit is $$ \lim_{t\to0^+}(-\cos t)\frac{\log(1+\sin t)}{\sin t} $$ and the second factor can be computed with $t=\arcsin u$: $$ \lim_{t\to0^+}\frac{\log(1+\sin t)}{\sin t} = \lim_{u\to0^+}\frac{\log(1+u)}{u} $$

(Note: log is the natural logarithm.)