Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.5 Exercises: 17

Answer

This function has no vertical asymptote.

Work Step by Step

$g(t)=\dfrac{t-1}{t^{2}+1}$ To find the vertical asymptote of this function, set its denominator equal to $0$ and solve for $t$: $t^{2}+1=0$ Take $1$ to the right side: $t^{2}=-1$ Take the square root of both sides: $\sqrt{t^{2}}=\sqrt{-1}$ $t=\sqrt{-1}$ Since $\sqrt{-1}$ is not a real number, this function has no vertical asymptote.
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