Limit x tending to 0 1 minus cos4theta whole divided by 1 minus cos 6 theta?

If x → 0, then : ( sin mx ) ~ ( mx ). (1)  
.. lim (x→0) [ ( 1 - cos 4x ) / ( 1 - cos 6x ) ] 
= lim (x→0) [ ( 2 sin² 2x ) / ( 2 sin² 3x ) ] dx 
= lim (x→0) { [ ( sin 2x )( sin 2x ) ] / [ ( sin 3x )( sin 3x ) ] } 
= lim (x→0) [ (2x) (2x) ] / [ (3x) (3x) ] from (1) 
= lim (x→0) ( 4 / 9 ) 
= 4/9 Ans.  
Happy To Help ! 

(1-cos(4x))/(1-cos(6x))
l'hopital's rule can be applied because it results in 0/0

take deritiive of top and bottom

(4sin(4x))/(6sin(6x))
if we evaluate
still 0/0
take derititive again

(-16cos(4x))/(-36(cos(6x))
iff we do x=0
-16/-36
16/36
4/9 is the limit

it approaches 4/9

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for

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