Q. 2 - Calculus

Question

Approximating limits: Use sequences of approximations to estimate the values of

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{4-x^2}{2-x} \\ \underset{z\to 3}{\lim }\frac{z^3-27}{z-3} \end{gathered}$$

Step-by-Step Solution

Answer

Limits values are following.

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{4-x^2}{2-x}=4 \\ \underset{z\to 3}{\lim }\frac{z^3-27}{z-3}=27 \end{gathered}$$

Step 1. Given information.

Given limits are the following.

$$\begin{gathered} f\left(x\right)=\underset{x\to 2}{\lim }\frac{4-x^2}{2-x} \\ f\left(x\right)=\underset{z\to 3}{\lim }\frac{z^3-27}{z-3} \end{gathered}$$

Step 2. Limit value of lim x → 2 4 - x 2 2 - x

Find the Limit value $\underset{x\to 2}{\lim }\frac{4-x^2}{2-x}.$

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{4-x^2}{2-x}=\underset{x\to 2}{\lim }\frac{2^2-x^2}{2-x} \\ =\underset{x\to 2}{\lim }\frac{\left(2-x\right)\left(2+x\right)}{2-x} \\ =\underset{x\to 2}{\lim }\left(x+2\right) \\ =2+2 \\ =4 \end{gathered}$$

So the value of $\underset{x\to 2}{\lim }\frac{4-x^2}{2-x}$is $4.$

Step 3. Limit value of lim z → 3 z 3 - 27 z - 3

Determine the limit value $\underset{z\to 3}{\lim }\frac{z^3-27}{z-3}.$

$$\begin{gathered} \underset{z\to 3}{\lim }\frac{z^3-27}{z-3}=\underset{z\to 3}{\lim }\frac{z^3-3^3}{z-3} \\ =\underset{z\to 3}{\lim }\frac{\left(z-3\right)\left(z^2+3x+3^2\right)}{z-3} \\ =\underset{z\to 3}{\lim }\left(z^2+3x+9\right) \\ =3^2+3\left(3\right)+9 \\ =27 \end{gathered}$$

So the value of $\underset{z\to 3}{\lim }\frac{z^3-27}{z-3}$is $27.$