evaluate the following limits: 1. lim x→3− 2/(x - 3)= 2. lim x→5 2/(x - 5)^6= 3. lim x→3+ 2/(x - 3)= 4. lim…

Explanation:

Step1: Analyze left - hand limit as $x\to3^{-}$

As $x\to3^{-}$, $x - 3\to0^{-}$. So, $\lim_{x\to3^{-}}\frac{2}{x - 3}=\frac{2}{0^{-}}=-\infty$.

Step2: Analyze limit as $x\to5$

As $x\to5$, $(x - 5)^6\to0^{+}$ (since the exponent is even). So, $\lim_{x\to5}\frac{2}{(x - 5)^6}=\frac{2}{0^{+}}=\infty$.

Step3: Analyze right - hand limit as $x\to3^{+}$

As $x\to3^{+}$, $x - 3\to0^{+}$. So, $\lim_{x\to3^{+}}\frac{2}{x - 3}=\frac{2}{0^{+}}=\infty$.

Step4: Analyze left - hand limit as $x\to - 7^{-}$

As $x\to - 7^{-}$, $x+7\to0^{-}$ and $x^{2}>0$. So, $\lim_{x\to - 7^{-}}\frac{1}{x^{2}(x + 7)}=\frac{1}{(+)(0^{-})}=-\infty$.

Answer:

1. $-\infty$
2. $\infty$
3. $\infty$
4. $-\infty$