Limits - Indeterminate FormsDate: 10/12/97 at 13:43:29 From: Mary Subject: Limits - Indeterminate Forms I cannot do the problem lim ((1/x) - (cot x)) x>0 I realize that this needs to be converted into the form 0/0 and then I must use L'Hopital's Rule; however I have tried and I cannot do this. I am having a similar problem with the question lim (1/x)(ln (7x+8)/(4x+8)) x>0 Any help you can give me would be greatly appreciated. Thank you. Date: 10/12/97 at 17:45:31 From: Doctor Anthony Subject: Re: Limits - Indeterminate Forms >lim ((1/x) - (cot x)) >x>0 1 cos(x) Write this as --- - ------ x sin(x) sin(x) - x.cos(x) ----------------- x.sin(x) If you expand sin(x) and cos(x) in series, you get x - x^3/3! + x^5/5! - ... - x[1 - x^2/2! + x^4/4! - ..] ------------------------------------------------------- x[x - x^3/3! + x^5/5! - ...] -x^3/3! + x^5/5! - ... + x^3/2! - x^5/4! + ... ----------------------------------------------- x^2 - x^4/3! + x^6/5! - ...... divide top and bottom by x^2 -x/3! + x^3/5! - .... + x/2! - x^3/4! + ... -------------------------------------------- 1 - x^2/3! + x^4/5! - ......... now if we let x -> 0 the top line -> 0 , but the bottom line -> 1, and so the final result -> 0/1 = 0 >lim (1/x)(ln (7x+8)/(4x+8)) >x>0 Again, series expansion is a good way to find the limit. 7x+8 3x ---- = 1 + ----- 4x+8 4x+8 and expand ln(1+y) as y - y^2/2 + y^3/3 - y^4/4 + ...... 3x 9x^2 27x^3 ln(1 + 3x/(4x+8)) = ------ - --------- + ---------- - ........ 4x+8 2(4x+8)^2 3(4x+8)^3 3 9x 27x^2 and so (1/x)ln(1 + 3x/(4x+8)) = ----- - --------- + -------- - .... 4x+8 2(4x+8)^2 3(4x+8)^3 Now let x -> 0 and this expression reduces to 3/8 So the limit as x -> 0 will be 3/8 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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