Q. 44 - Calculus

Question

Calculate each of the limits in Exercises $29-70$.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

Step 2 . The limit is calculated as below:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{x+x^2-2x^3}{x-x^2}=\underset{x\to 0}{\lim }\frac{x\left(1+x-2x^2\right)}{x(1-x)} \\ =\underset{x\to 0}{\lim }\frac{1+2x-x-2x^2}{(1-x)} \\ =\underset{x\to 0}{\lim }\frac{1(1+2x)-x(1+2x)}{(1-x)} \\ =\underset{x\to 0}{\lim }\frac{(1+2x)(1-x)}{(1-x)} \\ =\underset{x\to 0}{\lim }(1+2x) \end{gathered}$$

Step 3 . Let us use the sum rule in the obtained expression.

$$\begin{gathered} \underset{x\to 0}{\lim }(1+2x)=\underset{x\to 0}{\lim }\left(1\right)+\underset{x\to 0}{\lim }\left(2x\right) \\ =1+2\left(0\right) \\ =1+0 \\ =1 \end{gathered}$$