# Thread: trouble evaluating a limit

1. ## trouble evaluating a limit

lim y-(4*y^1/2)+3/y^2-1
x-1
I multiplied the numerator and denomination by the conjugate:
y-(4*y^1/2)+3 *y+(4*y^1/2)-3
y^2-1 *y+(4*y^1/2)-3

In the numerator, I get
y^2-16y-9

denomination gives me
y^2-1(y+(4*y^1/2)-3

I can factor the denominator a bit and i'm left with

y^2-16y-9/(y-1)(y+1)*(y+(4*y^1/2)-3

I can see that i should have got a factor of (y-1) in the numerator so that I could cancel the (y-1 ) in the denominator. I'm not sure what I am doing wrong. What did I do wrong and what am I missing?

2. $\frac{y - 4\sqrt{y} + 3}{y^2 - 1}$

let $\sqrt{y} = u$ ...

$\frac{u^2 - 4u + 3}{u^4 - 1}$

$\frac{(u - 3)(u - 1)}{(u^2 + 1)(u+1)(u-1)}$

$\frac{u-3}{(u^2 + 1)(u + 1)}$

back substitute $u = \sqrt{y}$

$\frac{\sqrt{y} - 3}{(y + 1)(\sqrt{y} + 1)}$

take the limit now?

3. Originally Posted by yoleven
lim y-(4*y^1/2)+3/y^2-1
x-1
I multiplied the numerator and denomination by the conjugate:
y-(4*y^1/2)+3 *y+(4*y^1/2)-3
y^2-1 *y+(4*y^1/2)-3

In the numerator, I get
y^2-16y-9

denomination gives me
y^2-1(y+(4*y^1/2)-3

I can factor the denominator a bit and i'm left with

y^2-16y-9/(y-1)(y+1)*(y+(4*y^1/2)-3

I can see that i should have got a factor of (y-1) in the numerator so that I could cancel the (y-1 ) in the denominator. I'm not sure what I am doing wrong. What did I do wrong and what am I missing?
Let's see if we can sort out your question by converting to LaTeX...

$\lim_{y\to\infty}\frac{y-4y^{\frac{1}{2}}+3}{y^2-1}$

I'm not sure if this is actually the equation you had in mind, or where the $x-1$ comes into play. However, if that is what you're trying to solve, then I can tell you that doing this:

$\lim_{y\to\infty}[\frac{y-4y^{\frac{1}{2}}+3}{y^2-1}][\frac{y+4y^{\frac{1}{2}}-3}{y+4y^{\frac{1}{2}}-3}]$

...won't help one bit, because it doesn't equal what you seem to think it does. Rather:

$\lim_{y\to\infty}[\frac{y-4y^{\frac{1}{2}}+3}{y^2-1}][\frac{y+4y^{\frac{1}{2}}-3}{y+4y^{\frac{1}{2}}-3}]=\lim_{y\to0}\frac{y^2-16y+24y^{\frac{1}{2}}-9}{y^3+4y^{\frac{5}{2}}-3y^2-y-4y^{\frac{1}{2}}+3}$

And obviously this is unhelpful. Actually, you can do this problem without performing any operations. Consider the original limit:

$\lim_{y\to\infty}\frac{y-4y^{\frac{1}{2}}+3}{y^2-1}$

You can see that the denominator is going to grow faster than the numerator, because it contains the highest power (2). So, as $y$ approaches infinity, the denominator is going to get larger and larger with respect to the numerator, which means the limit is zero. However, if you want to see the operations, just divide all the terms by $y$ to the highest power in any term (which, as before, is 2):

$\lim_{y\to\infty}\frac{y-4y^{\frac{1}{2}}+3}{y^2-1}=\lim_{y\to\infty}\frac{\frac{y}{y^2}-\frac{4y^{\frac{1}{2}}}{y^2}+\frac{3}{y^2}}{\frac{ y^2}{y^2}-\frac{1}{y^2}}$

And then simplify:

$\lim_{y\to\infty}\frac{\frac{1}{y}-\frac{4}{y^{\frac{3}{2}}}+\frac{3}{y^2}}{1-\frac{1}{y^2}}=\frac{0-0+0}{1-0}=0$

4. ## limit

the limit was as y goes to -1 not infinity but thanks.
I can definitely take the limit now.

5. Originally Posted by yoleven
the limit was as y goes to -1 not infinity but thanks.
I can definitely take the limit now.
I don't see how, since the domain is $0\leq y;y\neq 1$.

6. . . . quagmire . . . bet a dollar it's this:

$\lim_{y\to 1}\frac{y-4y^{1/2}+3}{y^2-1}=-1/2$ via L'Hospital's rule. May I recommend one of the finest rules to success in Mathematics: writing it as pretty as possible.

7. Sorry if my question was hard to decipher. The question was to look just like how skeeter put it in the original reply.
I don't know how to make my questions look clearer when I post them.
Sorry again for the ambiguity.

8. Originally Posted by yoleven
Sorry if my question was hard to decipher. The question was to look just like how skeeter put it in the original reply.
I don't know how to make my questions look clearer when I post them.
Sorry again for the ambiguity.
If one of the terms is $y^{\frac{1}{2}}$, or, $\sqrt{y}$, then the limit as $y$ approaches $-1$ doesn't exist, since $y$ doesn't exist for negative values.

9. Originally Posted by yoleven
Sorry if my question was hard to decipher. The question was to look just like how skeeter put it in the original reply.
I don't know how to make my questions look clearer when I post them.
Sorry again for the ambiguity.
Alright, guess that was crummy of me to post that. Sorry. I still don't know what the question was though. And writing neatly (as long as it's correct) is really a good way to be successful with math.