Another epsilon-delta proof

1. Oct 25, 2009

nietzsche

1. The problem statement, all variables and given/known data

Prove that

$$\lim_{x \to 3} \frac{1}{x-4} = -1$$

2. Relevant equations

3. The attempt at a solution

Given $$\varepsilon > 0$$, we want to find $$\delta$$ such that

\begin{align*} 0<|x-3|<\delta \Rightarrow |\frac{1}{x-4}+1| < \varepsilon \end{align*} \begin{align*} |\frac{1}{x-4}+1| &< \varepsilon\\ |\frac{x-3}{x-4}| &< \varepsilon \end{align*}

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

2. Oct 25, 2009

Staff: Mentor

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.

3. Oct 25, 2009

nietzsche

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

\begin{align*} \frac{1}{2} < |x-4| < \frac{3}{2}\\ 2 > |\frac{1}{x-4}| > \frac{2}{3} \end{align*}

and then plugging it into the epsilon formula

\begin{align*} |\frac{x-3}{x-4}| < \varepsilon\\ 2|x-3| < \varepsilon\\ |x-3| < \frac{\varepsilon}{2} \end{align*}

so we take

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

does it seem okay?

4. Oct 26, 2009

nietzsche

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

5. Oct 26, 2009

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)