Question

Use a graphing device to determine whether the limit exists. If the limit exists, estimate its value to two decimal places.

$\underset{x\to 1}{\lim }\frac{x^3+x^2+3x-5}{2x^2-5x+3}$

Short answer

Yes, the limit exists.

The value of limit of the given function is$-8.00$

Step-by-step solution

Step 1. Given information.

The limit of function here given is,

$\underset{x\to 1}{\lim }\frac{x^3+x^2+3x-5}{2x^2-5x+3}$

Step 2. Simplification.

$\begin{array}{c}\frac{x^3+x^2+3x-5}{2x^2-5x+3}:\frac{x^2+2x+5}{2x-3}\\ \underset{x\to 1}{\lim }\left(\frac{x^2+2x+5}{2x-3}\right)\end{array}$

Plug in the value $x=1$

$\begin{array}{c}=\frac{1^2+2\cdot 1+5}{2\cdot 1-3}\\ =-8\end{array}$

Step 3. Draw the graph.

Now, as a graphical verification use a graphing device.

Therefore, from the graph it is determined that$\underset{x\to 1}{\lim }\frac{x^3+x^2+3x-5}{2x^2-5x+3}=-8$