21. lim (cosec x -cotx)x-30
Let the function be,
f ( x ) = ( cos e c x − cot x )
We have to find the value of the function at limit x → 0 .
First we need to check the function by substituting the value.
So at x = 0 , the function is in ( ∞ − ∞ ) form.
We need to simplify the function in order to make it into simpler and standard form.
f ( x ) = cos e c x − cot x = 1 sin x − cos x sin x = ( 1 − cos x ) sin x
From the theorem of limits, we know that for any two functions f and g , such that both lim x → a f ( x ) and lim x → a g ( x ) exist, then
lim x → a f ( x ) g ( x ) = lim x → a f ( x ) l i m x → a g ( x ) (1)
According to the trigonometric theorem,
lim x → 0 sin x x = 1 (2)
On applying limits to the expression using equations (1) and (2), we get
lim x → 0 1 − c o s x sin x = lim x → 0 ( 1 − cos x ) lim x → 0 sin x
On multiplying and dividing the above expression with a variable x , we get
lim x → 0 ( 1 − cos x ) lim x → 0 sin x = lim x → 0 ( 1 − cos x x ) lim x → 0 ( sin x x )
From the formula of limits,
lim x → 0 ( 1 − cos x ) x = 0 (3)
So from equations (2) and (3), final expression will be:
lim x → 0 ( 1 − cos x x ) lim x → 0 ( sin x x ) = ( 0 1 ) = 0
Thus, the value of the given expression lim x → 0 ( cos e c x − cot x ) = 0
Let the function be,
We have to find the value of the function at limit
First we need to check the function by substituting the value.
So at
We need to simplify the function in order to make it into simpler and standard form.
From the theorem of limits, we know that for any two functions
According to the trigonometric theorem,
On applying limits to the expression using equations (1) and (2), we get
On multiplying and dividing the above expression with a variable
From the formula of limits,
So from equations (2) and (3), final expression will be:
Thus, the value of the given expression
