lim-x-pi-tan-x-1-cos-x-

Question Number 160609 by cortano last updated on 03/Dec/21

\lim_{x \to π} (\frac{\tan x}{1 + \cos x}) = ?

Answered by Ar Brandon last updated on 03/Dec/21

L = \lim_{x \to π} (\frac{\tan x}{1 + \cos x}), u = x - π

= \lim_{u \to 0} (\frac{\tan (u + π)}{1 + \cos (u + π)}) = \lim_{u \to 0} (\frac{\tan u}{1 - \cos u})

= \lim_{u \to 0} (\frac{\frac{u}{1 - \frac{u^{2}}{2}}}{1 - (1 - \frac{u^{2}}{2})}) = \lim_{u \to 0} (\frac{2 u}{2 - u^{2}} \cdot \frac{2}{u^{2}})

= \lim_{u \to 0} (\frac{4}{2 u - u^{3}}) \to \pm \infty

Answered by alephzero last updated on 03/Dec/21

\lim_{x \to π} (\frac{\tan x}{1 + \cos x}) = ?

T o b e h o n e s t, I^{'} m i n 6^{t h} g r a d e .

B u t I c a n t r y

\lim_{x \to π} \frac{\tan x}{1 + \cos x} = ?

\tan π = 0

\cos π = - 1

1 + \cos π = 0

T h i s e q u a t i o n m a y l o o k w e i r d,

b u t {l e t}^{'} s r e m e m b e r t h e r u l e s o f

a r i t h m e t i c .

\forall n {n \in R ∣ \frac{n}{n} = 1}

T h e n,

\lim_{x \to π} \frac{\tan x}{1 + \cos x} = \frac{\tan π}{1 + \cos π} = \frac{0}{0} = 1

O r, i f

\lim_{x \to 0} \frac{n}{x} \to \infty, w h e r e n \in R

t h e n

\lim_{x \to π} \frac{\tan x}{1 + \cos x} = \lim_{x \to 0} \frac{0}{x} = \infty

I f t h i s i s f a l s e, {d o n}^{'} t m i n d m e,

I {d o n}^{'} t k n o w L^{'} H {\hat{o p i t a l}}^{'} s r u l e,

a n d I^{'} m o n l y 12 y e a r s o l d .

Answered by tounghoungko last updated on 03/Dec/21

\lim_{x \to π} (\frac{2 \sin \frac{1}{2} x \cos \frac{1}{2} x}{2 \cos^{2} \frac{1}{2} x \cos x}) =

\lim_{x \to π} (\frac{\tan \frac{1}{2} x}{\cos x}) = - \infty

Answered by Mathspace last updated on 03/Dec/21

f (x) = \frac{t a n x}{1 + c o s x}

c h a n g e m e n t x - π = t g i v e

f (x) = f (t + π) = \frac{t a n (t + π)}{1 + c o s (t + π)} (t \to 0)

= \frac{t a n t}{1 - c o s t} \sim \frac{t}{\frac{t^{2}}{2}} = \frac{2}{t} \to \infty