Question

lim x approches to pai secx/1+tanx

Answer 1

Answer:

-1

Step-by-step explanation:

$\lim_{x \to \pi} \dfrac{ Sec\:x}{1 + Tan\:x}\\\\\text{Converting Sec x and Tan x to Sin and Cos forms we get,}\\\\\implies \lim_{x \to \pi} \dfrac{\dfrac{1}{Cos\:x}}{1 + \dfrac{Sin\:x}{Cos\:x}}\\\\\\\text{Taking LCM in the denominator we get,}\\\\\implies \lim_{x \to \pi} \dfrac{ \dfrac{1}{Cos\:x}}{ \dfrac{ Cos\:x + Sin\:x}{Cos\:x}}$

Denominator gets cancelled and we get,

$lim_{x \to \pi} \dfrac{1}{ Sin\:x + Cos\:x}$

Applying limits we get,

⇒ 1 / Sin π + Cos π

⇒ 1 / 0 + (-1)

⇒ 1 / -1 = -1

Hence the answer is -1.