Question

question attempt: 1 of unlimited
the model approximates that the canadian population will reach 40 million during the year 2027
correct answer
2028
part: 4 / 6
part 5 of 6
(e) what value will the term \\( \frac{9.6}{e^{0.02515t}} \\) approach as \\( t \to \infty \\)?
as \\( t \to \infty \\), the term \\( \frac{9.6}{e^{0.02515t}} \to \square \\)

Explanation:

Step1: Analyze the exponential term

As \( t
ightarrow\infty \), consider the behavior of \( e^{0.02515t} \). The exponential function \( y = e^{kt}\) with \(k = 0.02515>0\) grows without bound as \(t
ightarrow\infty\). That is, \(e^{0.02515t}
ightarrow\infty\) as \(t
ightarrow\infty\).

Step2: Evaluate the limit of the fraction

We have the fraction \(\frac{9.6}{e^{0.02515t}}\). Using the property of limits \(\lim_{t
ightarrow\infty}\frac{a}{f(t)}\) where \(a = 9.6\) (a constant) and \(f(t)=e^{0.02515t}
ightarrow\infty\) as \(t
ightarrow\infty\). By the rule \(\lim_{x
ightarrow\infty}\frac{c}{x}=0\) for a constant \(c\), we substitute \(x = e^{0.02515t}\) and \(c = 9.6\). So \(\lim_{t
ightarrow\infty}\frac{9.6}{e^{0.02515t}}=0\).

Answer:

\(0\)