Q. 28 - Calculus

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.

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

Step-by-Step Solution

Answer

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

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]

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