Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 12

Show that the function $\displaystyle h(t) = \frac{2t - 3t^2}{1 + t^3}$ is continuous at the given number $a = 1$ using the definition of continuity and the properties of limits.

By using properties of limit.

$\begin{equation} \begin{aligned} & \lim \limits_{x \to 1} \left( \frac{2t - 3t^2}{1 + t^3} \right) &&= \frac{2\lim \limits_{x \to 1} t - 3\lim \limits_{x \to 1} t^2}{\lim \limits_{x \to 1} 1 + \lim \limits_{x \to 1} t^3} && \text{Apply Difference, Sum and Quotient Law}\\ \\ & \phantom{x} && = \frac{2(1) - 3(1)^2}{1 + (1)^3} && \text{Substitute given value of $a$}\\ & \phantom{x} && = -\frac{1}{2} \end{aligned} \end{equation}$


By using the definition of continuity
$\lim \limits_{x \to a} f(x) = f(a)$





$\begin{equation} \begin{aligned} & \lim \limits_{x \to 1} \left( \frac{2t - 3t^2}{1 + t^3} \right) && = \frac{2(1)-3(1)^2}{1+(1)^3}\\ & \phantom {x} && = -\frac{1}{2} \end{aligned} \end{equation}$


Therefore, by applying either of the two, we have shown that the function is continuous at 1 and is equal to $\displaystyle -\frac{1}{2}$