How do I Find the Right Delta for an Epsilon-Delta Proof?

[nietzsche]

Homework Statement

Prove that

<br /> \lim_{x \to 3} \frac{1}{x-4} = -1<br />

Homework Equations

The Attempt at a Solution

Given \varepsilon &gt; 0, we want to find \delta such that

<br /> \begin{align*}<br /> 0&lt;|x-3|&lt;\delta \Rightarrow |\frac{1}{x-4}+1| &lt; \varepsilon<br /> \end{align*}<br /> \begin{align*}<br /> |\frac{1}{x-4}+1| &amp;&lt; \varepsilon\\<br /> |\frac{x-3}{x-4}| &amp;&lt; \varepsilon<br /> \end{align*}<br />

I always get stuck at this part! I'm never sure of how to proceed. Should I restrict the interval?

 

[Mark44]

You could could restrict |x - 3| to be less than .5, which means that 2.5 < x < 3.5, and -1.5 < x - 4 < -.5, so .5 < |x - 4| < 1.5.

 

[nietzsche]

so once we get to here, using \delta_1 = \frac{1}{2},

<br /> \begin{align*}<br /> \frac{1}{2} &lt; |x-4| &lt; \frac{3}{2}\\<br /> 2 &gt; |\frac{1}{x-4}| &gt; \frac{2}{3}<br /> \end{align*}<br />

and then plugging it into the epsilon formula

<br /> \begin{align*}<br /> |\frac{x-3}{x-4}| &lt; \varepsilon\\<br /> 2|x-3| &lt; \varepsilon\\<br /> |x-3| &lt; \frac{\varepsilon}{2}<br /> \end{align*}<br />

so we take

\delta = \text{min} (\frac{1}{2}, \frac{\varepsilon}{2})

does it seem okay?

 

[nietzsche]

sorry, i never got a response on this one, and I'm still not sure if it's right.

 

[emyt]

|(x-3)/(x-4)| < epsilon, |x-3| =/= 1, take delta = 1/2 --> - 1/2 < x-3 < 1/2, 5/2 < x < 3 7/2 --> -3/2 < x - 4 < -1/2 --> |x-4| > 1/2 --> |(x-3)/(x-4)| < 2|(x-3)| < epsilon


in short, delta = min(1/2 , epsilon/2)