Science:Math Exam Resources/Courses/MATH100 B/December 2023/Question 30(b)/Solution 1

In order to compute ${\textstyle \lim _{x\to \infty }S(x)}$ we first divide the numerator and denominator by ${\textstyle x}$: $$S(x)={\frac {x}{x\left(1+{\frac {k_{1}}{x}}\right)}}-{\frac {x}{x\left(1+{\frac {k_{2}}{x}}\right)}}={\frac {1}{1+{\frac {k_{1}}{x}}}}-{\frac {1}{1+{\frac {k_{2}}{x}}}}.$$ We can use the following theorem

If ${\textstyle r>0}$ is a rational number, then ${\textstyle \lim _{x\to \infty }{\frac {1}{x^{r}}}=0,}$

to compute

$$\lim _{x\to \infty }S(x)=\lim _{x\to \infty }\left({\frac {1}{1+{\frac {k_{1}}{x}}}}-{\frac {1}{1+{\frac {k_{2}}{x}}}}\right)={\frac {1}{1+\lim _{x\to \infty }\left({\frac {k_{1}}{x}}\right)}}-{\frac {1}{1+\lim _{x\to \infty }\left({\frac {k_{2}}{x}}\right)}}={\frac {1}{1+0}}-{\frac {1}{1+0}}=1-1=0.$$

Interpretation: As ${\textstyle x\to \infty }$, i.e., at very large doses of drug, the toxic effect dominates. Thus, ${\textstyle \lim _{x\to \infty }S(x)=0}$ means that, in this model, at extremely high doses, effectively no patients survive.