# Prove that when x tends to infinity, the limit of " (cot x)/x " does not exist.

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### 1 Answer

To show the limit of (cot x)/x as x-->infinity .

Solution

(cotx)/x = (cosx/sinx)/x = cosx/(xsinx).

Therefore Lt (cotx)/x = Lt.(cosx)/(xsinx) as x-->infinity.Now we consider 3 cases (i) x = npi , (ii) x = n(pi + a) and (iii) x = n(pi - a) where 0< a < pi, and n is a positive integer.

Case (i) x = npi , cosnpi = 1 or -1 , and sinnpi = 0. But (cos npi)/(sin npi) is indeterminate . So at x= npi , cotnpi is indeterminate. So at x= npi , Lt (cot npi)/npi is also indeterminate and does not exist . Both right limit and left limit are also indeterminates. So at x = npi , Lt n--> infinity (cot npi)/(npi) does ot exist.

case(ii)

At x = npi+ a , lt x--> inf (cotx)/x = Lt n--> inf {cot(npi+a)/(npi+a) = {cos(npi+a)/sin(pi+a) }/(npi+a) = (a finite number)/infinity = 0 for all 0 < a < pi.

case(iii) x= npi-a.

Lt x--> inf (cotx)/x = Lt n-->inf cot(npi-a)/(npi-a) = (A finite number)/ (infinity) = 0 for all 0 < a < pi.

Therefore Lt x--> infinity (cotx)/x does not approach a unique limit.

Hope this helps.