Question

# Determine the following limits. a. $\lim _{t \rightarrow-2^{+}} \frac{t^{3}-5 t^{2}+6 t}{t^{4}-4 t^{2}}$. b. $\lim _{t \rightarrow-2^{-}} \frac{t^{3}-5 t^{2}+6 t}{t^{4}-4 t^{2}}$. c. $\lim _{t \rightarrow-2} \frac{t^{3}-5 t^{2}+6 t}{t^{4}-4 t^{2}}$. d. $\lim _{t \rightarrow 2} \frac{t^{3}-5 t^{2}+6 t}{t^{4}-4 t^{2}}$.

Solution

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$\lim\limits_{t \to -2^+}\frac{t^3-5t^2+6t}{t^4-4t^2}$

Rewrite as:

$\lim\limits_{t \to -2^+}\frac{t^3-5t^2+6t}{t^4-4t^2}=\lim\limits_{t \to -2^+}\left(t^3-5t^2+6t\right)\cdot \lim\limits_{t \to -2^+}\frac{1}{t^4-4t^2}$

Observe:

\begin{align*} \lim\limits_{t \to -2^+} \left(t^3-5t^2+6t\right)&=-8-20-12\\ &=-40\\\\ \lim\limits_{t \to -2^+}\frac{1}{t^4-4t^2}&=\frac{1}{0^-}\\ &=-\infty \end{align*}

Finally,

\begin{align*} \lim\limits_{t \to -2^+}\frac{t^3-5t^2+6t}{t^4-4t^2}&=\lim\limits_{t \to -2^+}\left(t^3-5t^2+6t\right)\cdot \lim\limits_{t \to -2^+}\frac{1}{t^4-4t^2}\\ &=-40\cdot\left(- \infty\right)\\ &=\infty\end{align*}

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