evaluate the following limits. consider looking at the graph of the function situation. as necessary, enter…

evaluate the following limits. consider looking at the graph of the function situation. as necessary, enter oo for ∞ and -oo for -∞.\n(a) $lim_{x\rightarrow\frac{3}{2}^+}left(\frac{21x}{9 - 6x}\right)=$\n(b) $lim_{x\rightarrow\frac{3}{2}^-}left(\frac{21x}{9 - 6x}\right)=$\nquestion help: message instructor\nsubmit question jump to answer

evaluate the following limits. consider looking at the graph of the function situation. as necessary, enter oo for ∞ and -oo for -∞.\n(a) $lim_{x\rightarrow\frac{3}{2}^+}left(\frac{21x}{9 - 6x}\right)=$\n(b) $lim_{x\rightarrow\frac{3}{2}^-}left(\frac{21x}{9 - 6x}\right)=$\nquestion help: message instructor\nsubmit question jump to answer

Answer

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) $\infty$