Kyle R.

asked • 09/25/16

Limit at infinity with cot and tan

limx-->positive infinity tan ( x/x^2+1) 
limx--> negative infinity cot ( pix/4x+1) 
This is so confusing for me, i would lice a detailed step by step. 

1 Expert Answer


Lee H. answered • 09/25/16

5 (17)

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Kyle R.

Can I just do (x/x)*(1/x+1) to get 0? Im not sure why you have to divode by x^2. 


Lee H.

Sorry, I interpreted the question as  tan ( x/(x^2+1)). If however it actually is tan ( (x/x^2)+1) Then yes you can just divide out an x from the fraction, or factor out x/x from the expression as you suggest. Then
limx-->positive infinity x/x^2+1 = limx-->positive infinity (x/x)*(1/x + 1) = 1*1 = 1, not 0.
So last step becomes tan(1) = 1.557408.  But: it does not make too much sense to write (x/x^2) + 1. Why not write 1/x + 1 in the first place? So please double check the original question to make sure it's (x/x^2) + 1 and not x/(x^2+1). - Lee


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