Q. 36 - Calculus

Question

Write each limit in Exercises 19–42 as a formal statement involving $\delta ,\epsilon ,N, \mathrm{and}/\mathrm{or} M,$ and sketch a graph that illustrates the roles of these constants. In the last six exercises you may take function to be any appropriate graph.

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{2+h}}-{\displaystyle \frac{1}{2}}}{h}=-\frac{1}{4}$

Step-by-Step Solution

Answer

For all $\epsilon >0$, there exists some $\delta >0$ such that if,

$x\in \left(-\delta ,0\right)\cup \left(0,\delta \right), \mathrm{then} \frac{{\displaystyle \frac{1}{2+h}}-{\displaystyle \frac{1}{2}}}{h}\in \left(-\frac{1}{4}-\epsilon ,-\frac{1}{4}+\epsilon \right)$.

The graph is 

Step 1. Given information.

The given expression is $\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{2+h}-\frac{1}{2}}}{h}=-\frac{1}{4}$.

Step 2. Explanation.

From the given expression, we have, $c=0, L=-\frac{1}{4}$.

The limit expression can be written as a formal statement as below,

For all $\epsilon >0$, there exists some $\delta >0$ such that if,

$x\in \left(-\delta ,0\right)\cup \left(0,\delta \right), \mathrm{then} \frac{{\displaystyle \frac{1}{2+h}-\frac{1}{2}}}{h}\in \left(-\frac{1}{4}-\epsilon ,-\frac{1}{4}+\epsilon \right)$.

Step 3. Graph the expression.

The graph illustrating the roles of the constants is shown as below: