find-the-value-of-lim-1-cosx-tan-2x-x-

Question Number 69174 by Aditya789 last updated on 21/Sep/19

f i n d t h e v a l u e o f

l i m \frac{1 + c o s x}{{t a n}^{2 x}}

x \hat{-} ⇈

Commented by Henri Boucatchou last updated on 21/Sep/19

N o t e t h a t t h e r e a r e s o m e f a i l u r e s i n y o u r e d i t i o n

Commented by Rasheed.Sindhi last updated on 21/Sep/19

D o y o u m e a n :

\underset{x \to π}{l i m} \frac{1 + c o s x}{t a n (2 x)}

O r y o u m e a n :

\underset{x \to π}{l i m} \frac{1 + c o s x}{{t a n}^{2} x}

?

Commented by mathmax by abdo last updated on 21/Sep/19

i f y o u m e a n {l i m}_{x \to π} \frac{1 + c o s x}{t a n (2 x)} w e d o t h e c h a n g e m e n t x = π + t \Rightarrow

\frac{1 + c o s x}{t a n (2 x)} = \frac{1 + c o s (π + t)}{t a n (2 π + 2 t)} = \frac{1 - c o s t}{t a n (2 t)} a n d x \to π \Rightarrow t \to 0 s o

\frac{1 - c o s t}{t a n (2 t)} \sim \frac{\frac{t^{2}}{2}}{2 t} = \frac{t}{4} \to 0 (t \to 0) s o {l i m}_{x \to π} \frac{1 + c o s x}{t a n (2 x)} = 0

i f y o u m e a n {l i m}_{x \to π} \frac{1 + c o s x}{{t a n}^{2} x} c h a n g . x = π + t g i v e

\frac{1 + c o s x}{{t a n}^{2} x} = \frac{1 - c o s t}{{t a n}^{2} (t)} \sim \frac{t^{2}}{2} \times \frac{1}{t^{2}} = \frac{1}{2} \Rightarrow {l i m}_{x \to π} \frac{1 + c o s x}{{t a n}^{2} x} = \frac{1}{2}