Question

evaluate the following limits. consider looking at the graph of the function situation. as necessary, enter oo for ∞ and -oo for -∞.
(a) $lim_{x
ightarrow\frac{3}{2}^+}left(\frac{21x}{9 - 6x}
ight)=$
(b) $lim_{x
ightarrow\frac{3}{2}^-}left(\frac{21x}{9 - 6x}
ight)=$
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Explanation:

Step1: Analyze the limit of the rational - function

We have the function $y = \frac{21x}{9 - 6x}$. We can rewrite it as $y=\frac{21x}{-6x + 9}$. When finding the limit as $x\to\frac{3}{2}^{-}$ (left - hand limit), we consider the behavior of the function near $x=\frac{3}{2}$.

Step2: Use the properties of limits

As $x\to\frac{3}{2}^{-}$, the numerator $21x\to21\times\frac{3}{2}=\frac{63}{2}$ (a positive value), and the denominator $9 - 6x\to0^{+}$ (a small positive number). Since $\lim_{x\to\frac{3}{2}^{-}}\frac{21x}{9 - 6x}=\frac{\text{positive}}{\text{small positive}}=\infty$.
As $x\to\frac{3}{2}^{+}$, the numerator $21x\to21\times\frac{3}{2}=\frac{63}{2}$ (a positive value), and the denominator $9 - 6x\to0^{-}$ (a small negative number). So, $\lim_{x\to\frac{3}{2}^{+}}\frac{21x}{9 - 6x}=\frac{\text{positive}}{\text{small negative}}=-\infty$.

Answer:

(a) $-\infty$
(b) $\infty$