Chapter 5, Problem 3RWDT

To determine

An example of functions $f$ and $g$ such that the given conditions are satisfied.

Answer to Problem 3RWDT

It has been determined that for $f\left(x\right)=\frac{x^2+c^2}{x-c}$ and $g\left(x\right)=\frac{2xc}{x-c}$ , $\underset{x\to c}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=0$ but $\underset{x\to c}{\lim }f\left(x\right)$ and $\underset{x\to c}{\lim }g\left(x\right)$ do not exist.

Explanation of Solution

Given:

  $\underset{x\to c}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=0$ but $\underset{x\to c}{\lim }f\left(x\right)$ and $\underset{x\to c}{\lim }g\left(x\right)$ do not exist.

Concept used:

If the denominator of a function is zero at some point, then the value of the function at that point does not exist.

Calculation:

Let $f\left(x\right)=\frac{x^2+c^2}{x-c}$ and $g\left(x\right)=\frac{2xc}{x-c}$ .

Clearly, $\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to c}{\lim }\frac{x^2+c^2}{x-c}$ and $\underset{x\to c}{\lim }g\left(x\right)=\underset{x\to c}{\lim }\frac{2xc}{x-c}$ do not exist because the denominator becomes zero at the limit point.

Now, $f\left(x\right)-g\left(x\right)=\frac{x^2+c^2}{x-c}-\frac{2xc}{x-c}$ .

Simplifying,

  $f\left(x\right)-g\left(x\right)=\frac{x^2-2xc+c^2}{x-c}$

On further simplification,

  $f\left(x\right)-g\left(x\right)=\frac{{\left(x-c\right)}^2}{x-c}$

Continuing simplification,

  $f\left(x\right)-g\left(x\right)=x-c$

Then, it follows that,

  $\underset{x\to c}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=\underset{x\to c}{\lim }\left(x-c\right)=0$

So, for $f\left(x\right)=\frac{x^2+c^2}{x-c}$ and $g\left(x\right)=\frac{2xc}{x-c}$ , $\underset{x\to c}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=0$ but $\underset{x\to c}{\lim }f\left(x\right)$ and $\underset{x\to c}{\lim }g\left(x\right)$ do not exist.

Conclusion:

It has been determined that for $f\left(x\right)=\frac{x^2+c^2}{x-c}$ and $g\left(x\right)=\frac{2xc}{x-c}$ , $\underset{x\to c}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=0$ but $\underset{x\to c}{\lim }f\left(x\right)$ and $\underset{x\to c}{\lim }g\left(x\right)$ do not exist.