Question

evaluate the following limits:

1. $lim_{x

ightarrow5^{-}}\frac{2}{(x - 5)^{3}}$

1. $lim_{x

ightarrow3^{-}}\frac{2}{x - 3}$

1. $lim_{x

ightarrow0}\frac{1}{x^{2}(x + 7)}$

1. $lim_{x

ightarrow - 7^{-}}\frac{1}{x^{2}(x + 7)}$

Explanation:

Step1: Analyze the first - limit

As $x\to5^{-}$, let $t=x - 5$, then $t\to0^{-}$. The function becomes $\lim_{t\to0^{-}}\frac{2}{t^{3}}$. Since $t\to0^{-}$, $t^{3}\to0^{-}$ and $\frac{2}{t^{3}}\to-\infty$.

Step2: Analyze the second - limit

As $x\to3^{-}$, let $u=x - 3$, then $u\to0^{-}$. The function becomes $\lim_{u\to0^{-}}\frac{2}{u}$. Since $u\to0^{-}$, $\frac{2}{u}\to-\infty$.

Step3: Analyze the third - limit

As $x\to0$, we have $\lim_{x\to0}\frac{1}{x^{2}(x + 7)}$. Since $\lim_{x\to0}x^{2}=0$ and $\lim_{x\to0}(x + 7)=7$, and $x^{2}\gt0$ for $x
eq0$, then $\lim_{x\to0}\frac{1}{x^{2}(x + 7)}=\infty$.

Step4: Analyze the fourth - limit

As $x\to - 7^{-}$, let $v=x + 7$, then $v\to0^{-}$ and $x=v - 7$. The function becomes $\lim_{v\to0^{-}}\frac{1}{(v - 7)^{2}v}$. Since $(v - 7)^{2}\gt0$ for $v\to0^{-}$ and $v\to0^{-}$, then $\lim_{v\to0^{-}}\frac{1}{(v - 7)^{2}v}=-\infty$.

Answer:

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