Question

01) Verifique se existe indeterminações. Se existir, retire as indeterminações, fatorando os polinômios ou dividindo os polinômios por um fator comum. Calcule os limites.
$\lim_{t \to -1} \frac{3t^2 + t -2}{4t^2 + 5t+1}$

Answer 1

Olá

$\displaystyle \lim_{t \to -1}~ \frac{3t^2+t-2}{4t^2+5t+1} ~=~\frac{3(-1)^2+(-1)-2}{4(-1)^2+5(-1)+1}~=~ \frac{3-3}{4-4} ~=~ \frac{0}{0} \\ \\ \\ \text{Divide ambos os polinomios por t+1 (raiz do limite)} \\ \\ \\ 3t^2+t-2~~~~~ ~\div~~~~~ ~~|\underline{t+1} \\ \underline{-3t^2-3t} ~~ ~~~~~ ~~ ~~~~~~~~ ~\boxed{3t-2 } \\ 0-2-2 \\ \underline{~+2t+2} \\ 0~~~~ ~~0$

$\displaystyle 4t^2+5t+1~~~~~ ~\div~~~~~ ~~|\underline{t+1} \\ \underline{-4t^2-4t}~~~~ ~~~~~~ ~~~~~~ ~~~\boxed{4t+1} \\ 0+t+1 \\ \underline{-t-1 ~~~~} \\ 0~~~0$

$\text{Substituindo no limite} \\ \\ \\ \displaystyle \lim_{t \to -1}~ \frac{(x+1)\cdot(3t-2)}{(x+1)\cdot(4t+1)} \\ \\ \\ \text{Simplifica os termos em comum (x+1)} \\ \\ \\ \lim_{t \to -1}~~~ ~~\frac{\diagup\!\!\!\!\!\!\!\!\!\!\!(t+1)\cdot(3t-2)}{\diagup\!\!\!\!\!\!\!\!\!\!\!(t+1)\cdot(4t+1)} \\ \\ \\ \lim_{t \to -1} ~\frac{3t-2}{4t+1} ~=~ \frac{3(-1)-2}{4(-1)+1} ~=~ \frac{-3-2}{-4+1} ~=~ \frac{-5}{-3}~=~\boxed{\boxed{ \frac{5}{3} }}$