How do you use the definition of continuity and the properties of limits to show that the function h(t)=(2t-3t^2)/(1+t^3) is continuous at the given number a=1?

Jan 14, 2018

Explanation:

A function $f$ is continuous at number $a$ if and only if ${\lim}_{x \rightarrow a} f \left(x\right) = f \left(a\right)$.

(In order to be equal, both must exist.)

In ths problem, we need to show that, for $h \left(t\right) = \frac{2 t - 3 {t}^{2}}{1 + {t}^{3}}$ we get ${\lim}_{t \rightarrow 1} h \left(t\right) = h \left(1\right)$

The tools we have to work with are the properties of limits. (These are quite standard, but if yours differ little, I hope you can still get the idea.

${\lim}_{t \rightarrow 1} h \left(t\right) = {\lim}_{t \rightarrow 1} \frac{2 t - 3 {t}^{2}}{1 + {t}^{3}}$

$= \frac{{\lim}_{t \rightarrow 1} \left(2 t - 3 {t}^{2}\right)}{{\lim}_{t \rightarrow 1} \left(1 + {t}^{3}\right)}$

if both limits exist and the limit in the denominator is not $0$ by the quotient property of limits

$= \frac{{\lim}_{t \rightarrow 1} \left(2 t\right) - {\lim}_{t \rightarrow 1} \left(3 {t}^{2}\right)}{{\lim}_{t \rightarrow 1} \left(1\right) + {\lim}_{t \rightarrow 1} \left({t}^{3}\right)}$

if all limits exist and the denominator is not $0$ by the sum and difference properties of limits

$= \frac{2 {\lim}_{t \rightarrow 1} \left(t\right) - 3 {\lim}_{t \rightarrow 1} \left({t}^{2}\right)}{{\lim}_{t \rightarrow 1} \left(1\right) + {\lim}_{t \rightarrow 1} \left({t}^{3}\right)}$

if all limits exist and the denominator is not $0$ by the constant multiple property of limits

$= \frac{2 {\lim}_{t \rightarrow 1} \left(t\right) - 3 {\left({\lim}_{t \rightarrow 1} \left(t\right)\right)}^{2}}{{\lim}_{t \rightarrow 1} \left(1\right) + {\left({\lim}_{t \rightarrow 1} \left(t\right)\right)}^{3}}$

if all limits exist and the denominator is not $0$ by the power property of limits

$= \frac{2 \left(1\right) - 3 {\left(1\right)}^{2}}{\left(1\right) + {\left(1\right)}^{3}}$

By evaluating the limit of a constant and the limit of the identity function. Note that the limits do exist and the denominator is not $0$. So

${\lim}_{t \rightarrow 1} h \left(t\right) = \frac{2 \left(1\right) - 3 {\left(1\right)}^{2}}{\left(1\right) + {\left(1\right)}^{3}} = h \left(1\right)$