# Help evaluating limits

• Jul 11th 2007, 05:21 PM
starswept
Help evaluating limits
I'm having difficulty evaluating the following limits:

lim x->4

2-√x
_______
3-√2x+1

and

lim x->0

2^2x - 2^x
___________
2^x-1

I'd really appreciate any help with these. I know I need to multiply by the conjugate radical in the first one, but after that I'm stuck. With the second, I'm really not sure what to do. Thanks in advance :)
• Jul 11th 2007, 05:26 PM
galactus
Quote:

$\lim_{x\rightarrow{0}}\frac{2^{2x}-2^{x}}{2^{x}-1}$

I'd really appreciate any help with these. I know I need to multiply by the conjugate radical in the first one, but after that I'm stuck. With the second, I'm really not sure what to do. Thanks in advance :)
Factor out 2^x: $\frac{2^{x}(2^{x}-1)}{2^{x}-1}=2^{x}$

This simplifies down to $\lim_{x\rightarrow{0}}2^{x}$

Now, the limits easy?.
• Jul 11th 2007, 05:31 PM
starswept
Quote:

Originally Posted by galactus
Factor out 2^x: $\frac{2^{x}(2^{x}-1)}{2^{x}-1}=2^{x}$

This simplifies down to $\lim_{x\rightarrow{0}}2^{x}$

Now, the limits easy?.

That's what I thought... but wouldn't that make the limit equal to 1, because 2^0 = 1? The back of my textbook has the answer as 16. Did I do something wrong?
• Jul 11th 2007, 05:37 PM
galactus
No, you're correct. The limit is 1. If that is the problem, then the book is wrong.
• Jul 11th 2007, 06:18 PM
Jhevon
Quote:

Originally Posted by starswept
I'm having difficulty evaluating the following limits:

lim x->4

2-√x
_______
3-√2x+1

you really should use brackets after the square root symbol so we know everything that is being rooted. if you look carefully, your problem looks like $\frac {2 - \sqrt {x}}{3 - \sqrt {2x} + 1}$ when i know it's supposed to be $\frac {2 - \sqrt {x}}{3 - \sqrt {2x + 1}}$. in this case it is easy to see what the problem should be, but that won't always be the case. use parenthesis to clarify, it will save you and the person responding to you a lot of frustration.

Hint. Rationalize the denominator. some algebraic manipulation would be needed after this. Think: "difference of two squares"
• Jul 11th 2007, 06:33 PM
starswept
Quote:

Originally Posted by Jhevon
you really should use brackets after the square root symbol so we know everything that is being rooted. if you look carefully, your problem looks like $\frac {2 - \sqrt {x}}{3 - \sqrt {2x} + 1}$ when i know it's supposed to be $\frac {2 - \sqrt {x}}{3 - \sqrt {2x + 1}}$. in this case it is easy to see what the problem should be, but that won't always be the case. use parenthesis to clarify, it will save you and the person responding to you a lot of frustration.

Hint. Rationalize the denominator. some algebraic manipulation would be needed after this. Think: "difference of two squares"

I'm sorry, I know it looks confusing. I'll use brackets from now on. I'm hoping as soon as I find a few minutes where I'm not at school or doing homework, I can learn how to insert equations/type math nicely so that at least when I come on here to ask inane questions it's in proper format ;)

I rationalized to:

6 + 2(√2x+1) - 3(√x) - (√x)(√2x+1)
______________________________

8-2x

My issue is that I don't know how to multiply those last two terms: (√x)( √2x+1). Does the sign cancel out?
• Jul 11th 2007, 06:38 PM
Jhevon
Quote:

Originally Posted by starswept
I'm sorry, I know it looks confusing. I'll use brackets from now on. I'm hoping as soon as I find a few minutes where I'm not at school or doing homework, I can learn how to insert equations/type math nicely so that at least when I come on here to ask inane questions it's in proper format ;)

I rationalized to:

6 + 2(√2x+1) - 3(√x) - (√x)(√2x+1)
______________________________

8-2x

My issue is that I don't know how to multiply those last two terms: (√x)( √2x+1). Does the sign cancel out?

remember, we are concerned with the bottom, we worry about simplifying the top after we make sure the bottom does not go to zero. in the above formula, the bottom still goes to zero. more manipulation is necessary. don't multiply out the top, keep it in factored form, and then, think "difference of two squares"

see here to get a copy of the LaTex tutorial and learn how to type "pretty math"
• Jul 11th 2007, 07:04 PM
starswept
Quote:

Originally Posted by Jhevon
remember, we are concerned with the bottom, we worry about simplifying the top after we make sure the bottom does not go to zero. in the above formula, the bottom still goes to zero. more manipulation is necessary. don't multiply out the top, keep it in factored form, and then, think "difference of two squares"

see here to get a copy of the LaTex tutorial and learn how to type "pretty math"

Thank you! I finally got it. I completely overlooked the fact that 4-1x is a difference of squares. And thanks for the link as well :)
• Jul 11th 2007, 07:06 PM
Jhevon
Quote:

Originally Posted by starswept
Thank you! I finally got it. I completely overlooked the fact that 4-1x is a difference of squares. And thanks for the link as well :)

you're most welcome. glad i could be of assistance. good luck in your class