Question

In Problems 7– 42, find each limit algebraically

$\underset{x\to -1}{\lim }\left(8x^5-7x^3+8x^2+x-4\right)$

Short answer

The value of the limit is$2$

Step-by-step solution

Step 1. Given Information

The given limit is$\underset{x\to -1}{\lim }\left(8x^5-7x^3+8x^2+x-4\right)$

We need to find the value of the limit.

Step 2. Expanding the limit

We know,

$$\begin{gathered} \underset{x\to c}{\lim }\left[f\right(x)+g(x\left)\right]=\underset{x\to c}{\lim }f\left(x\right)+\underset{x\to c}{\lim }g\left(x\right) \\ \underset{x\to c}{\lim }\left[f\right(x)-g(x\left)\right]=\underset{x\to c}{\lim }f\left(x\right)-\underset{x\to c}{\lim }g\left(x\right) \end{gathered}$$

Therefore,

$$\begin{gathered} \underset{x\to -1}{\lim }\left(8x^5-7x^3+8x^2+x-4\right)=\underset{x\to -1}{\lim }8x^5-\underset{x\to -1}{\lim }7x^3+\underset{x\to -1}{\lim }8x^2+\underset{x\to -1}{\lim }x-\underset{x\to -1}{\lim }4 \\ =8·\underset{x\to -1}{\lim }{x}^5-7·\underset{x\to -1}{\lim }{x}^3+8·\underset{x\to -1}{\lim }{x}^2+\underset{x\to -1}{\lim }x-\underset{x\to -1}{\lim }4 \end{gathered}$$

Step 3. Substitution of limit

The simplified expression will be

$$\begin{gathered} =8·\underset{x\to -1}{\lim }{x}^5-7·\underset{x\to -1}{\lim }{x}^3+8·\underset{x\to -1}{\lim }{x}^2+\underset{x\to -1}{\lim }x-\underset{x\to -1}{\lim }4 \\ =8·(-1{)}^5-7·(-1{)}^3+8·(-1{)}^2+(-1)-4 \\ =-8+7+8-1-4 \\ =2 \end{gathered}$$