Find the vertical asymptotes of the graph of the function

AI Thread Summary
The vertical asymptotes of the function tan(x)/x occur at x = π(2n + 1)/2, where n is an integer. There is no vertical asymptote at x = 0 because the limit of the function as x approaches 0 is 1. The initial solution provided was confirmed as correct by other participants in the discussion. The clarification about the behavior at x = 0 was appreciated. This reinforces the understanding of vertical asymptotes in relation to the function's limits.
Barbados_Slim
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Homework Statement


Find the vertical asymptotes of the graph of the function. (Use n as an arbitrary integer)
\frac{tanx}{x}


Homework Equations


N/A


The Attempt at a Solution


I believe the answer is x=\frac{\pi(2n+1)}{2}. I would just like somebody to confirm or deny this. Thanks in advance.
 
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If you are right, why is there no vert. asymptote at x = 0 ?
 
Is it because the limit as x approaches 0 is 1?
 
Correct !

So your answer in post #1 of this thread is right !
 
Thank you for your help.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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