Question

suppose the function $g$ satisfies $-32x - 2x^{2}-137 < g(x)ightarrow - 8}g(x)$. first evaluate: $lim_{x
ightarrow - 8}-32x - 2x^{2}-137$. next evaluate: $lim_{x
ightarrow - 8}x^{2}+16x + 55$. therefore, by the squeeze theorem, $lim_{x
ightarrow - 8}g(x)=$

Explanation:

Step1: Evaluate $\lim_{x

ightarrow - 8}(-32x - 2x^{2}-137)$
Substitute $x = - 8$ into $-32x - 2x^{2}-137$.

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Step2: Evaluate $\lim_{x

ightarrow - 8}(x^{2}+16x + 55)$
Substitute $x=-8$ into $x^{2}+16x + 55$.

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Step3: Apply Squeeze Theorem

Since $\lim_{x
ightarrow - 8}(-32x - 2x^{2}-137)=\lim_{x
ightarrow - 8}(x^{2}+16x + 55)=-9$ and $-32x - 2x^{2}-137<g(x)<x^{2}+16x + 55$, by the Squeeze Theorem, $\lim_{x
ightarrow - 8}g(x)=-9$.

Answer:

The value of $\lim_{x
ightarrow - 8}(-32x - 2x^{2}-137)$ is $-9$.
The value of $\lim_{x
ightarrow - 8}(x^{2}+16x + 55)$ is $-9$.
The value of $\lim_{x
ightarrow - 8}g(x)$ is $-9$.