Without calculus, how to determine the behaviour of the rational function $f(t)= \frac {30t} {t^2 + 2}$ for small values of $t$ ( i.e. near $0$).

Question

Source : Stewart's Precalculus,

The question reads like this :

The concentration ( in Mg/L) of a drug injected to a patient is a function of time that follows the formula :

$$f(t)= \frac {30t} {t^2 + 2} \space \space , \space \space \text {with} \space t\geq 0 $$

(a) Draw the graph of the drug concentration.

(b) What eventually happens to the drug concentration.

Part (b) can be solved by alledging that the degree of the above polynomial is less than the degree of the bottom one, which means that $\lim _{t\rightarrow \infty} \frac {30t} {t^2 + 2} = 0$.

My question deals with part (a), and in particular with the behaviour of the function for small values of $x$.

A graphing calculator shows me that when $t$ is small, the graph of function $f$ is very close to the graph of $y= 30t/2 = 15t$.

But if I had no such tool at my disposal, how could I know this , without using calculus?

The only argument I can find is that, when $t$ is less than $1$ , $t^2$ is less than $t$, in such a way that the $t^2$ term in the denominator can be neglected. But that does not sound very rigorous.

Answer 1

You have$$f(t)=\frac{30t}{t^2+2}=15t\times\frac1{1+t^2/2}$$and, when $t$ is close to $0$, $\frac1{1+t^2/2}$ behaves as $1$.

Answer 2

You have answered yourself. For $|t|<\epsilon$ we have that $\left|\dfrac{30t}{t^2+2}-15t\right|=\left|\dfrac{15t^3}{t^2+2}\right|<\dfrac{15\epsilon}2$