True or False:
x → 0 lim n x sin ( m x ) = n m
If m , n are non-zero reals then the above equation holds true.
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That big line equation is not true. What are you missing from the central terms?
No need for Mr. L' Hopital's rule since x → 0 implies mx → 0. Hence limit simplifies to: n m ( x → 0 lim ( m x ) sin ( m x ) ) = n m
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You must prove x → 0 lim m x sin ( m x ) = 1
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1< x sin x <cosx and apply Squeeze Theorem lim x → 0 1 < lim x → 0 x sin x < lim x → 0 cos x ⇒ x → 0 lim x sin x = 1
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@Rishabh Jain – Yes correct. For the same purpose I applied L' Hopital's (Cause I learnt it recently).
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@Nihar Mahajan – No problem... (+1) for a Neatly written solution.. : D
@Rishabh Jain – 1 is not less than x s i n x when x → 0 ...
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@Guillermo Templado – That's why.. Since lim x → 0 1 = lim x → 0 cos x = 1 Hence by Squeeze theorem we also claim lim x → 0 x sin x = 1 .
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@Rishabh Jain – ok, look at this
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@Guillermo Templado – sin x < x < tan x when x > 0 and x belongs to neighborhood of 0. Now , we are going to divide for sin x . Then, sin x sin x < sin x x < cos x 1 and the we can apply squezze theorem when x → 0 + .(I can give you another arguments when x < 0) sin x is a vertical segment and x is an angle (we are talking in radians at the circrcle of radius 1 (geometry)) for that sin x < x when x > 0 . I can give you one analytical demonstration of this too, if you desire it
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@Guillermo Templado – Sure.. Please go ahead.. :-)
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@Rishabh Jain – Ok, take f ( x ) = x − sin x this function fullfies f ( 0 ) = 0 ⇒ 0 = s i n 0 and f ′ ( x ) = 1 − cos x > 0 when x ∈ ( 0 , 4 π ) so this function is strictly increasing in this interval , and this means x > sin x when x ∈ ( 0 , 4 π ) ,you can apply the mean value theorem (Lagrange theorem) or also take Taylor polynomyal of sin x = x − 3 ! x 3 ... graph x - sinx
When x < 0, then we have tan x < x < sin x in a neighborhood of 0 (this time x < 0) and we can also apply the squeeze theorem when x → 0 −
Just use definition of derivative of Sin function at 0 witch lead to Cos(0) = 1 .
In response to Challenge Master: Lol , I was being silly. I have corrected it now.
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Note that as x → 0 , the expression x sin x becomes 0 0 . Hence we can use L' Hopital's Rule : x → 0 lim x sin x = x → 0 lim 1 cos x = cos ( 0 ) = 1
Hence we have:
x → 0 lim n x sin ( m x ) = x → 0 lim m n x m sin ( m x ) = x → 0 lim n m × x → 0 lim m x sin ( m x ) = n m
Hence , the result is indeed true!