help with limit definitions

**seamstress** · February 28th 2011, 11:27 PM

I'm going crazy trying to understand this problem, like it's literally driving me mad. Please someone help explain this to me?

Use the definition of the limit to show that
$\displaystyle\lim_{x \to -1}(\frac{x+5}{2x+3}) = 4$

So I said "Given any $\epsilon > 0$ , there exists a $\delta > 0$ such that if $0 < \mid x+1 \mid < \delta$ , then $\mid \frac{x+5}{2x+3} - 4 \mid < \epsilon$

So I simplified $\mid \frac{x+5}{2x+3} - 4 \mid$ to $7 \mid \frac{x+1}{2x+3} \mid$ , and I get stuck.

The biggest issue I have is that the "answer" (but it really doesn't give you a proof) is in the back of the book, and I'm pulling my hair out trying to understand it.
This is what they say:

If $\mid x+1 \mid < \frac{1}{4}$ , then $\mid \frac{x+5}{2x+3}-4 \mid < 7 \mid \frac{x+1}{2x+3} \mid$
Q: How did they say that $\mid x+1 \mid < \frac{1}{4}$ . Why were they able to just say that it is less than 1/4?

Then this is where I really get confused:

$7 \mid\frac{x+1}{2x+3}\mid < 14\mid x+1\mid$
Q: Where did they get this from? I understand that it is less than that, but I don't get how they got that. Like if I were doing a different problem, would I just pull that out of thin air?

Then they say we may take $\delta := inf \{\frac{1}{4}, \frac{\epsilon}{14}\}$

Q: I kind of understand this, so that it will always be less than $\epsilon$ , but I'm again not sure where they got this?

If anyone could help me, that would be awesome. I really want so badly to understand how to do this, I just can't do it alone!

Thank you.

**CaptainBlack** · March 1st 2011, 12:00 AM

Originally Posted by seamstress

I'm going crazy trying to understand this problem, like it's literally driving me mad. Please someone help explain this to me?

Use the definition of the limit to show that
$\displaystyle\lim_{x \to -1}(\frac{x+5}{2x+3}) = 4$

So I said "Given any $\epsilon > 0$ , there exists a $\delta > 0$ such that if $0 < \mid x+1 \mid < \delta$ , then $\mid \frac{x+5}{2x+3} - 4 \mid < \epsilon$

So I simplified $\mid \frac{x+5}{2x+3} - 4 \mid$ to $7 \mid \frac{x+1}{2x+3} \mid$ , and I get stuck.

The biggest issue I have is that the "answer" (but it really doesn't give you a proof) is in the back of the book, and I'm pulling my hair out trying to understand it.
This is what they say:

If $\mid x+1 \mid < \frac{1}{4}$ , then $\mid \frac{x+5}{2x+3}-4 \mid < 7 \mid \frac{x+1}{2x+3} \mid$
Q: How did they say that $\mid x+1 \mid < \frac{1}{4}$ . Why were they able to just say that it is less than 1/4?

Since $$$x$ is going to be going to $$$-1$ we can consider only those $$$x$ 's closer to $$$-1$ than $$$1/4$ - that is all this is saying

Then this is where I really get confused:

$7 \mid\frac{x+1}{2x+3}\mid < 14\mid x+1\mid$
Q: Where did they get this from? I understand that it is less than that, but I don't get how they got that. Like if I were doing a different problem, would I just pull that out of thin air?

You want to get a bound that gets rid of the $2x+3$ term in the denominator, so you find the minimum value of $|2x+3|$ when $|x+1|<1/4$ and use that in place of the $2x+3$ to give an upper bound.

Then they say we may take $\delta := inf \{\frac{1}{4}, \frac{\epsilon}{14}\}$

Q: I kind of understand this, so that it will always be less than $\epsilon$ , but I'm again not sure where they got this?

If anyone could help me, that would be awesome. I really want so badly to understand how to do this, I just can't do it alone!

Thank you.

There is no unique way of proceeding with these things, what you do is you play around with vaules to find ones that work. You have:

$\mid \frac{x+5}{2x+3} - 4 \mid=7 \mid \frac{x+1}{2x+3} \mid$

so now you need to find a $\delta>\left|\frac{7}{2x+3}\right|$ when $$$x$ is close to $$$-1$ .

CB

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