the increase in a persons body temperature t(t), above 98.6°f, can be modeled by the function t(t) = 4t /…

the increase in a persons body temperature t(t), above 98.6°f, can be modeled by the function t(t) = 4t / (t² + 1), where t represents time elapsed. what is the meaning of the horizontal asymptote for this function? the horizontal asymptote of y = 0 means that the persons temperature will approach 98.6°f as time elapses. the horizontal asymptote of y = 0 means that the persons temperature will approach 0°f as time elapses. the horizontal asymptote of y = 4 means that the persons temperature will approach 102.6°f as time elapses. the horizontal asymptote of y = 4 means that the persons temperature will approach 4°f as time elapses. the horizontal asymptote of y = 4 means that the persons temperature will approach 4°f as time elapses.
Answer
Explanation:
Step1: Analizar la función
Tenemos la función $T(t)=\frac{4t}{t^{2}+1}$. Para encontrar el asintótico horizontal, consideramos el límite cuando $t\to\pm\infty$.
Step2: Calcular el límite
Dividimos numerador y denominador por $t^{2}$: $\lim_{t\to\pm\infty}\frac{4t}{t^{2}+1}=\lim_{t\to\pm\infty}\frac{\frac{4t}{t^{2}}}{\frac{t^{2}}{t^{2}}+\frac{1}{t^{2}}}=\lim_{t\to\pm\infty}\frac{\frac{4}{t}}{1 + \frac{1}{t^{2}}}$. Como $\lim_{t\to\pm\infty}\frac{4}{t}=0$ y $\lim_{t\to\pm\infty}\frac{1}{t^{2}} = 0$, entonces $\lim_{t\to\pm\infty}\frac{\frac{4}{t}}{1+\frac{1}{t^{2}}}=0$. El valor base de la temperatura corporal es $98.6^{\circ}F$. El asintótico horizontal $y = 0$ significa que la temperatura adicional sobre $98.6^{\circ}F$ se aproxima a $0$, es decir, la temperatura del individuo se aproxima a $98.6^{\circ}F$ a medida que transcurre el tiempo.
Answer:
The horizontal asymptote of y = 0 means that the person's temperature will approach 98.6°F as time elapses.