Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\lim_{x\to0}\frac{x^2+2x}{x^2-3x} = - \frac{2}{3}$$

Given that $\epsilon = 0.01$, what is the largest possible $\delta$ according to the epsilon-delta definition of the limit?

Here's what I've done by myself, but I'm getting stuck:

$$\left|\frac{x^2+2x}{x^2-3x}+\frac{2}{3}\right|<\epsilon\ \text{ if }0 <x<\delta$$ $$\left|\frac{3x^2+6x+2x^2-6x}{3x^2-9x}\right|<\epsilon\ \text{ if }0 <x<\delta$$ $$\left|\frac{5x^2}{3x^2-9x}\right|<\epsilon\ \text{ if }0 <x<\delta$$ $$\left|\frac{5x}{3x-9}\right|<\epsilon\ \text{ if }0 <x<\delta$$

I could split the absolute value into two separate ones and multiply the denominator up to the other side, but that wouldn't get me any closer to isolating x. Even plugging in 0.01 for $\epsilon$ doesn't seem to help. What am I doing wrong?

share|improve this question
1  
"Near" $0$, you should have $3x-9 < 0$. So the fraction can be written as $\frac{5|x|}{9-3x}$. Now, split into 2 cases depending on the sign of $x$. –  Srivatsan Sep 3 '11 at 1:52

1 Answer 1

up vote 4 down vote accepted

I could split the absolute value into two separate ones and multiply the denominator up to the other side, but that wouldn't get me any closer to isolating x.

Sure does. When $x$ is small and negative, the fraction is positive, so:

$$\frac{5x}{3x-9}<\epsilon$$ $$5x > (3x-9)\epsilon$$ $$5x > 3\epsilon x - 9\epsilon$$ $$(5-3\epsilon)x > -9\epsilon$$ $$x > \frac{-9\epsilon}{5-3\epsilon}$$

where the inequality sign was flipped because $3x-9$ must be negative under this assumption. The case for positive $x$ is much the same (and I'll leave it to you to do the details).

share|improve this answer
1  
@Anonymous Now, splitting into two cases two thresholds: $\delta_{-} = \frac{9\epsilon}{5-3\epsilon}$, and a $\delta_{+} = \ldots$(exercise). The meaning is that whenever $-\delta_{-} < x < +\delta_{+}$, we have $|f(x) + 2/3| < \epsilon$. Now, if you want a single $\delta$ that works for both signs, then the answer is $\ldots$. (Fill in the blank.) –  Srivatsan Sep 3 '11 at 2:43
    
@Srivatsan Narayanan For $\delta_{+}$ I get $\delta = \frac{9 \epsilon}{5 - 3\epsilon}$, how can the two of them be equal? Do we take $\delta =$ max $\{\delta_{-}, \delta_{+}\}$? –  user38268 Sep 3 '11 at 6:10
1  
@D, who says they have to be equal? The question just asked for the largest $\delta$ that satisfies both branches. Suppose that instead of the expressions here we knew that the $\epsilon$ was satisfied for $-5>x>7$. Then setting $\delta=max(5,7)=7$ wouldn't work, because, say, $x=-6$ then satisfies $\delta$ but not $\epsilon$. –  Henning Makholm Sep 3 '11 at 11:36
    
@D B Lim And I suspect that $\delta_{+}$ is not $\frac{9\epsilon}{5-3\epsilon}$, but something else. Please check your calculation. –  Srivatsan Sep 3 '11 at 13:10
    
Oh, I see, I probably misunderstood D B's comment. FWIW, I get $9\epsilon/(5+3\epsilon)$ on the positive side. –  Henning Makholm Sep 3 '11 at 19:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.