Question

find the value of lim x--> 0 ( cosec x -cot x).

Answer 1

$\bold{\lim_{x \to \ 0} {cosecx -cotx}}$
First of all check which form of limit exist here,
put x = 0
cosec0 - cot0 = ∞ - ∞ , this is the form of limit
So firstly , we have to resolve the expression.
= $\bold{\lim_{x \to \ 0} {\frac{1}{sinx}-\frac{cosx}{sinx}}}$
= $\bold{\lim_{x \to \ 0} {\frac{1 - cosx}{sinx}}}$
We know, sin2Ф = 2sinФ.cosФ
And 1 - cosA = 2sin²A/2, use these here,
= $\bold{\lim_{x \to \ 0} {\frac{2sin^2(x/2)}{2sin(x/2)cos(x/2)}}}$
= $\bold{\lim_{x \to \ 0} {\frac{sin(x/2)}{cos(x/2)}}}$
= $\bold{\lim_{x \to \ 0} {tanx/2}}}$
Now, put x = 0
= Tan(0/2) = 0

Hence, answer is 0

Answer 2

HELLO DEAR,

$\lim_{x \to 0} (cosecx - cot x) \lim_{x \to 0} \frac{1}{sinx} - \frac{cosx}{sinx} \lim_{x \to 0} [\frac{(1 - cosx)}{sinx} ]$
∴ [ (1 - cosx) = (2sin²x/2)]

$\lim_{x \to 0} \frac{2 sin^{2}x }{2sin \frac{x}{2} *cos \frac{x}{2} }$
∴ [ sin2x = 2sinxcosx , sinx = 2sinx/2 × cosx/2 ]

$\lim_{x \to 0} \frac{2sin \frac{x}{2} }{cos \frac{x}{2} } \lim_{x \to 0} 2* \tan \frac{x}{2} Put x =0 we get, 2*tan0/2 = 0$


I HOPE ITS HELP YOU DEAR,
THANKS