finding deltas algebraically

**squizygirl07** · September 5th 2012, 05:02 AM

i am confused how to do this problem....
f(x) = 1/x
L = -1
c = -1
epsilon = .1

i got .9 for p and 1.1 for q, which makes the solutions for delta the same.
thanks, emma

**Prove It** · September 5th 2012, 05:55 AM

Originally Posted by squizygirl07

i am confused how to do this problem....
f(x) = 1/x
L = -1
c = -1
epsilon = .1

i got .9 for p and 1.1 for q, which makes the solutions for delta the same.
thanks, emma

I assume you're trying to show that $\displaystyle \begin{align*} \lim_{x \to -1} \frac{1}{x} = -1 \end{align*}$ . To do this, you need to show $\displaystyle \begin{align*} 0 < |x - (-1)| < \delta \implies \left| \frac{1}{x} - (-1) \right| < \epsilon \end{align*}$ , in other words, show $\displaystyle \begin{align*} 0 < |x + 1| < \delta \implies \left| \frac{1}{x} + 1 \right| < \epsilon \end{align*}$ .

Working on the last inequality, we have

$\displaystyle \begin{align*} \left|\frac{1}{x} + 1 \right| &< \epsilon \\ \left| \frac{x + 1}{x} \right| &< \epsilon \\ \frac{|x + 1|}{|x|} &< \epsilon \\ |x + 1| &< |x| \epsilon \end{align*}$

So we need a lower bound for $\displaystyle \begin{align*} |x| \end{align*}$ . Remembering that we are letting x get very close to -1, so that the distance between them is small, we can choose a value that will make that distance small (and we're going to make it smaller anyway). It doesn't matter what value we choose, as long as it is sufficiently close. So say we let $\displaystyle \begin{align*} |x + 1| < \frac{3}{2} \end{align*}$ . That means

$\displaystyle \begin{align*} -\frac{3}{2} < x + 1 &< \frac{3}{2} \\ -\frac{5}{2} < x &< \frac{1}{2} \\ |x| &< \frac{1}{2} \end{align*}$

Therefore we can say $\displaystyle \begin{align*} |x + 1| < \frac{\epsilon}{2} \end{align*}$ .

So an appropriate $\displaystyle \begin{align*} \delta \end{align*}$ is $\displaystyle \begin{align*} \min \left\{ \frac{3}{2}, \frac{\epsilon}{2} \right\} \end{align*}$ .

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