Un quesito sugli asintoti verticali
Data una funzione fratta, ci chiediamo: è sempre vero
che, se xo è uno zero del denominatore, la retta x= xo è
un asintoto verticale per la funzione? La risposta non è sempre affermativa. Infatti la funzione
f(x)=
sin(x)
x−π
MathType@MTEF@5@5@+=
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ha nel
punto
x=π
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uno
zero per il denominatore, ma l'equazione
x=π
MathType@MTEF@5@5@+=
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non
rappresenta un asintoto verticale per la funzione data. Per provare questa
affermazione basterà calcolare il limite:
lim
x→π
sin(x)
x−π
MathType@MTEF@5@5@+=
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. Tale limite si presenta nella forma di
indecisione 0/0 che occorre sciogliere. Ponendo:
y=x−π
MathType@MTEF@5@5@+=
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, segue
che
x=y+π
MathType@MTEF@5@5@+=
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e
quindi, al tendere di x a
π
MathType@MTEF@5@5@+=
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, y
tenderà a zero. Perciò
lim
x→π
sin(x)
x−π
MathType@MTEF@5@5@+=
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=
lim
y→0
sin(y+π)
y
MathType@MTEF@5@5@+=
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. Applicando la formula di addizione del seno:
sin(y+π)=sin(y)cos(π)+cos(y)sin(π)=sin(y)(−1)+cos(y)(0)=−sin(y)
MathType@MTEF@5@5@+=
feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn
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. Pertanto,
lim
y→0
sin(y+π)
y
MathType@MTEF@5@5@+=
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=
lim
y→0
−
sin(y)
y
=−1
MathType@MTEF@5@5@+=
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. Quindi la funzione data non ammette l’asintoto
verticale
x=π
MathType@MTEF@5@5@+=
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.