Q. 7 - Calculus

Question

Find functions f and g and a real number c such that $\left(\underset{x\to c}{\lim }f\left(x\right)\right)\left(\underset{x\to c}{\lim }g\left(x\right)\right)\ne \underset{x\to c}{\lim }\left(f\right(x\left)g\right(x\left)\right)$. Does this example contradict the product rule for limits? Why or why not?

Step-by-Step Solution

Answer

The function $f\left(x\right)=x$ and $g\left(x\right)=\frac{1}{x}$ and a real number $c=0$ such that $\left(\underset{x\to c}{\lim }f\left(x\right)\right)\left(\underset{x\to c}{\lim }g\left(x\right)\right)\ne \underset{x\to c}{\lim }\left(f\right(x\left)g\right(x\left)\right)$ and contradicts product rule for limits.

Step 1. Given information.

An example is to be written for the condition  $\left(\underset{x\to c}{\lim }f\left(x\right)\right)\left(\underset{x\to c}{\lim }g\left(x\right)\right)\ne \underset{x\to c}{\lim }\left(f\right(x\left)g\right(x\left)\right)$

Step 2. Example for the given condition.

Let $f\left(x\right)=x$ and $g\left(x\right)=\frac{1}{x}$ and a real number $c=0$ such that limit is $x\to 0$.

We have

$$\begin{gathered} \left(\underset{x\to 0}{\lim }f\left(x\right)\right)\left(\underset{x\to 0}{\lim }g\left(x\right)\right)=\left(\underset{x\to 0}{\lim }x\right)\left(\underset{x\to 0}{\lim }\frac{1}{x}\right) \\ =0\left(\frac{1}{0}\right) \\ =\mathrm{undefined} \end{gathered}$$

and

$$\begin{gathered} \underset{x\to 0}{\lim }\left(f\right(x\left)g\right(x\left)\right) \\ =\underset{x\to 0}{\lim }(x\times \frac{1}{x}) \\ =\underset{x\to 0}{\lim }\left(1\right) \\ =1 \end{gathered}$$

So, $\left(\underset{x\to c}{\lim }f\left(x\right)\right)\left(\underset{x\to c}{\lim }g\left(x\right)\right)\ne \underset{x\to c}{\lim }\left(f\right(x\left)g\right(x\left)\right)$ and this example contradict the product rule for limits which states that $\left(\underset{x\to c}{\lim }f\left(x\right)\right)\left(\underset{x\to c}{\lim }g\left(x\right)\right)=\underset{x\to c}{\lim }\left(f\right(x\left)g\right(x\left)\right)$.