# Look for a value of delta limit

1. Mar 3, 2005

### Benny

Q. Find the limit $$\mathop {\lim }\limits_{x \to 2} \left( {\frac{{2x - 1}}{{x + 1}}} \right)$$ and verify the result by $$\varepsilon - \delta$$ argument.

I got the limit as 1 so I began to look for a value of delta.

Required: $$\left| {\frac{{2x - 1}}{{x + 1}} - 1} \right| < \varepsilon$$ whenever $$0 < \left| {x - 2} \right| < \delta$$.

$$\left| {\frac{{x - 2}}{{x + 1}}} \right| < \varepsilon$$ whenever $$0 < \left| {x - 2} \right| < \delta$$.

I get stuck at this point. Even after I use division on the LHS I still don't get anywhere. Perhaps I need to consider values of x near 2 and deduce something from that? I dunno. Could someone please help me out? Any help would be great thanks.

2. Mar 3, 2005

### Justin Lazear

Note that |x - 2| = |x + 1 - 3| <= |x + 1| + |-3| <=> |x - 2| - 3 <= |x + 1| by the Triangle Inequality.

--J

3. Mar 3, 2005

### Benny

Thanks for your help so far. Using the inequality I get:

$$\left| {\frac{{x + 2}}{{x + 1}}} \right| \le \frac{{\left| {x + 2} \right|}}{{\left| {x + 2} \right| - 3}} \le \left| {x + 2} \right|$$

I then choose delta = epsilon. Is that step valid?

4. Mar 3, 2005

### Justin Lazear

No. I lead you a bit astray, actually. You'll note that your step is not valid, at, for instance, x = 1.1.

Also, if you're making your denominator larger, it's actually making the total smaller, so the <= sign is not preserved by the first substitution.

You'll find the other end of the triangle inequality much more useful.

|x - 2| = |x + 1 - 3| >= |x + 1| - |-3| <=> |x + 1| <= |x - 2| + 3.

--J

Last edited: Mar 3, 2005
5. Mar 3, 2005

### Benny

Ok thanks for the help. I'll give it another try.