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Question Number 213841 by mnjuly1970 last updated on 18/Nov/24

Find the vertical asymptots of , f(x)= tan((( π)/(2x + 2)) ) in [ 0 , 4 ] −−−−−−−−−−−−−

$$ \\ $$$$\:\:{Find}\:{the}\:{vertical}\:{asymptots} \\ $$$$\: \\ $$$$\:\:{of}\:\:,\:\:\:{f}\left({x}\right)=\:\mathrm{tan}\left(\frac{\:\pi}{\mathrm{2}{x}\:+\:\mathrm{2}}\:\right)\:\:{in}\: \\ $$$$\: \\ $$$$\:\:\:\:\:\left[\:\mathrm{0}\:\:,\:\:\:\mathrm{4}\:\right] \\ $$$$\:−−−−−−−−−−−−− \\ $$$$ \\ $$

Answered by efronzo1 last updated on 18/Nov/24

Let x= c be a vertical asymptot lim_(x→c) tan ((π/(2x+2))) = ∞ it follow that (π/(2x+2)) = (π/2) or (π/(2x+2)) = ((3π)/2) ⇒2x+2 = 2 ; x= 0 ⇒2x+2 = (2/3) ; x<0

$$\:\:\mathrm{Let}\:{x}=\:{c}\:\:\mathrm{be}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{asymptot}\: \\ $$$$\:\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2x}+\mathrm{2}}\right)\:=\:\infty\: \\ $$$$\:\:\mathrm{it}\:\mathrm{follow}\:\mathrm{that}\:\frac{\pi}{\mathrm{2x}+\mathrm{2}}\:=\:\frac{\pi}{\mathrm{2}}\:\mathrm{or}\:\frac{\pi}{\mathrm{2x}+\mathrm{2}}\:=\:\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$\:\:\Rightarrow\mathrm{2x}+\mathrm{2}\:=\:\mathrm{2}\:;\:\mathrm{x}=\:\mathrm{0} \\ $$$$\:\:\Rightarrow\mathrm{2x}+\mathrm{2}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:;\:\mathrm{x}<\mathrm{0}\: \\ $$

Commented by mnjuly1970 last updated on 18/Nov/24

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