Q. 6 - Calculus

Question

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

Step-by-Step Solution

Answer

The functions $f\left(x\right)=\frac{1}{x}$ and $g\left(x\right)=-\frac{1}{x}$ and a real number $c=0$ such that $\underset{x\to c}{\lim }f\left(x\right)+\underset{x\to c}{\lim }g\left(x\right)\ne \underset{x\to c}{\lim }\left(f\right(x)+g(x\left)\right)$ and contradict the sum rule for limits.

Step 1. Given information.

Given a condition $\underset{x\to c}{\lim }f\left(x\right)+\underset{x\to c}{\lim }g\left(x\right)\ne \underset{x\to c}{\lim }\left(f\right(x)+g(x\left)\right)$.

Step 2. Example for the given condition.

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

We have

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

and

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

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