Question

evaluate the following limits. if needed, enter oo for \\( \infty \\) and -oo for \\( -\infty \\).
(a) \\( \lim _{x \
ightarrow \infty}\left(29 x^{2}+27 x^{3}\
ight)= \\)
(b) \\( \lim _{x \
ightarrow-\infty}\left(29 x^{2}+27 x^{3}\
ight)= \\)
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Explanation:

Step1: Factor out the highest - degree term

For the function \(y = 29x^{2}+27x^{3}\), factor out \(x^{3}\). So \(29x^{2}+27x^{3}=x^{3}(27 + \frac{29}{x})\)

Step2: Evaluate the limit as \(x\to\infty\)

We know that \(\lim_{x
ightarrow\infty}\frac{1}{x}=0\). Then \(\lim_{x
ightarrow\infty}(29x^{2}+27x^{3})=\lim_{x
ightarrow\infty}x^{3}(27+\frac{29}{x})\)
Since \(\lim_{x
ightarrow\infty}x^{3}=\infty\) and \(\lim_{x
ightarrow\infty}(27+\frac{29}{x}) = 27\), by the property of limits \(\lim_{x
ightarrow a}(f(x)\cdot g(x))=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)\) (when both limits exist), we have \(\lim_{x
ightarrow\infty}(29x^{2}+27x^{3})=\infty\)

Step3: Evaluate the limit as \(x\to-\infty\)

We know that \(\lim_{x
ightarrow-\infty}\frac{1}{x}=0\). Then \(\lim_{x
ightarrow-\infty}(29x^{2}+27x^{3})=\lim_{x
ightarrow-\infty}x^{3}(27+\frac{29}{x})\)
Since \(\lim_{x
ightarrow-\infty}x^{3}=-\infty\) and \(\lim_{x
ightarrow-\infty}(27+\frac{29}{x}) = 27\), by the property of limits \(\lim_{x
ightarrow a}(f(x)\cdot g(x))=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)\) (when both limits exist), we have \(\lim_{x
ightarrow-\infty}(29x^{2}+27x^{3})=-\infty\)

Answer:

(a) \(oo\)
(b) \(-oo\)