Solución

Apartado a

Usa la ecuación 3.3 y sustituye el valor t=3t = 3 en los dos términos de la ecuación:
lim⁡t→3r(t)=lim⁡t→3[(t2−3t+4)i+(4t+3)j]=[lim⁡t→3(t2−3t+4)]i+[lim⁡t→3(4t+3)]j=4i+15j\begin{aligned} \lim\limits_{t \to 3}\bold{r}(t) &= \lim\limits_{t \to 3}[(t^2−3t+4)\bold{i}+(4t+3)\bold{j}]\\ &= \bigg[\lim\limits_{t \to 3}(t^2−3t+4)\bigg]\bold{i} + \bigg[\lim\limits_{t \to 3}(4t+3)\bigg]\bold{j}\\ &= 4\bold{i}+15\bold{j} \end{aligned}

Apartado b

Usa la ecuación 3.4 y sustituya el valor t=3t = 3 en las expresiones de la ecuación:
lim⁡t→3r(t)=lim⁡t→3(2t−4t+1i+tt2+1j+(4t−3)k)=[lim⁡t→3(2t−4t+1)]i+[lim⁡t→3(tt2+1)]j+[lim⁡t→3(4t−3)]k=12i+310j+9k\begin{aligned} \lim\limits_{t \to 3}\bold{r}(t) &= \lim\limits_{t \to 3}\bigg(\frac{2t−4}{t+1}\bold{i}+\frac{t}{t^2+1}\bold{j}+(4t−3)\bold{k}\bigg)\\ &= \bigg[\lim\limits_{t \to 3}\bigg(\frac{2t−4}{t+1}\bigg)\bigg]\bold{i} + \bigg[\lim\limits_{t \to 3}\bigg(\frac{t}{t^2+1}\bigg)\bigg]\bold{j} + \bigg[\lim\limits_{t \to 3}(4t−3)\bigg]\bold{k}\\ &= \frac{1}{2}\bold{i} + \frac{3}{10}\bold{j}+ 9\bold{k} \end{aligned}