assignment 2: problem 27\n(1 point)\nevaluate the following limits:\n1. $lim_{x\rightarrow5^{-}}\frac{2}{(x…

assignment 2: problem 27\n(1 point)\nevaluate the following limits:\n1. $lim_{x\rightarrow5^{-}}\frac{2}{(x - 5)^{3}}=square$\n2. $lim_{x\rightarrow5}\frac{2}{(x - 5)^{6}}=square$\n3. $lim_{x\rightarrow - 7^{-}}\frac{1}{x^{2}(x + 7)}=square$\n4. $lim_{x\rightarrow3^{+}}\frac{2}{x - 3}=square$\nnote: you can earn partial credit on this problem.

assignment 2: problem 27\n(1 point)\nevaluate the following limits:\n1. $lim_{x\rightarrow5^{-}}\frac{2}{(x - 5)^{3}}=square$\n2. $lim_{x\rightarrow5}\frac{2}{(x - 5)^{6}}=square$\n3. $lim_{x\rightarrow - 7^{-}}\frac{1}{x^{2}(x + 7)}=square$\n4. $lim_{x\rightarrow3^{+}}\frac{2}{x - 3}=square$\nnote: you can earn partial credit on this problem.

Answer

Explanation:

Step1: Analyze left - hand limit as $x\to5^{-}$

As $x\to5^{-}$, $(x - 5)\to0^{-}$. So, $\lim_{x\to5^{-}}\frac{2}{(x - 5)^{3}}=-\infty$.

Step2: Analyze limit as $x\to5$

As $x\to5$, $(x - 5)\to0$. Since the power is even, $\lim_{x\to5}\frac{2}{(x - 5)^{6}}=\infty$.

Step3: Analyze left - hand limit as $x\to - 7^{-}$

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

Step4: Analyze right - hand limit as $x\to3^{+}$

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

Answer:

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