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 $\displaystyle -\frac{1}{2}$