Question

Calculate the limits in Exercises $25-28$, using only the continuity of linear and power functions and the limit rules. Cite each limit rule that you apply.

localid="1648227587052" $\underset{x\to 0}{\lim }\frac{3}{2x^2-4x+1}$.

Short answer

The value of $\underset{x\to 0}{\lim }\frac{3}{2x^2-4x+1}$is,$3.$

Step-by-step solution

Step 1. Given information

$\underset{x\to 0}{\lim }\frac{3}{2x^2-4x+1}$.

Step 2. The limit is calculated as below,

$\underset{x\to 0}{\lim }\frac{3}{2x^2-4x+1}=\frac{\underset{x\to 0}{\lim }3}{\underset{x\to 0}{\lim }(2x^2-4x+1)}$[Quotient rule]

$=\frac{\underset{x\to 0}{\lim }3}{\underset{x\to 0}{\lim }\left(2x^2\right)-\underset{x\to 0}{\lim }4x+\underset{x\to 0}{\lim }\left(1\right)}$[Sum rule]

$$\begin{gathered} =\frac{3}{(2·0)-(4·0)+1} \\ =\frac{3}{1} \\ =3 \end{gathered}$$