Question

In Problems 7– 42, find each limit algebraically

$\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)$

Short answer

The value of the limit$\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)=-1$

Step-by-step solution

Step 1. Given Information

The given limit function is$\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)$

We need to find the value of the limit.

Step 2. Expanding the limit

We know,

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

Therefore,

$\underset{x\to 1}{\lim }\left(5x^4-3x^2+6x-9\right)=\underset{x\to 1}{\lim }5x^4-\underset{x\to 1}{\lim }3x^2+\underset{x\to 1}{\lim }6x-\underset{x\to 1}{\lim }9$

Step 3. Value of the limit

The simplified expression is

$$\begin{gathered} \underset{x\to 1}{\lim }5x^4-\underset{x\to 1}{\lim }3x^2+\underset{x\to 1}{\lim }6x-\underset{x\to 1}{\lim }9 \\ =\{5\times {\left(1\right)}^4\}-\left\{3{\left(1\right)}^2\right\}\}+6(1)-9 \\ =5-3+6-9 \\ =-1 \end{gathered}$$