Limit as X Goes to Infinity

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hopelynnwelch

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I need to find the limit as X goes to infinity of (3x^2-1)/(sqrt(5x^4+6))

What I did was make a table and sub in values of 10, 100, 1,000, and 10,000 and I see that as x goes to infinity the function goes to 1.34.

I was curious if I could manipulate the algebraically to come to a conclusion as well but I am not sure.

I was thinking maybe multiply the entire function to get rid on the root in the denominator but I don't know if that really helps at all... Any help appreciated.
 
I'm not sure if this is quite what you're looking for, but it's how I see it. Look at the highest exponent values for your numerator and denominator. In the numerator, you essentially have X1 and in the denominator, you essentially have X1/2. Logically, if you plug in increasing values of X, the numerator will increase more quickly than the denominator resulting in a higher number, which eventually reaches infinity.
 
hopelynnwelch said:
I need to find the limit as X goes to infinity of (3x^2-1)/(sqrt(5x^4+6))

What I did was make a table and sub in values of 10, 100, 1,000, and 10,000 and I see that as x goes to infinity the function goes to 1.34.

I was curious if I could manipulate the algebraically to come to a conclusion as well but I am not sure.

I was thinking maybe multiply the entire function to get rid on the root in the denominator but I don't know if that really helps at all... Any help appreciated.
Click to expand...
the limit as X goes to infinity of (3x^2-1)/(sqrt(5x^4+6))=
the limit as X goes to infinity of (3x^2)/(sqrt(5x^4)). Lets see what you can do with this.
 
murdazman13 said:
I'm not sure if this is quite what you're looking for, but it's how I see it. Look at the highest exponent values for your numerator and denominator. In the numerator, you essentially have X1 and in the denominator, you essentially have X1/2. Logically, if you plug in increasing values of X, the numerator will increase more quickly than the denominator resulting in a higher number, which eventually reaches infinity.
Click to expand...
Where do you see X1
Maybe you have (3x^2-1)^1 but not x^1. Same for the denominator.
Consider this: x^3/ sqrt(x^100). This reduces to x^3/x^50 = 1/x^47 which goes to 0 as x goes to infinity.
 
When I look at it as (3x^2)/(sqrt(5x^4)) I still get 1.34 for values of 10, 100, and 1000 etc... Right? I suck at these...
 
OHHHH WAIT! I am not getting closer to 1.34... and am just slowly and slowly getting further from 1.34. It's just super slow. So it is infinity... Right?
 
Or maybe I am still wrong and the answer is - INFINITY? I should graph this in google and check lol
 
I completely misread the OP. Wow, that's embarrassing. After rereading, i found that 3X2/(5X4)1/2 simplifies to 3/51/2. Does that help any?

Sorry for any confusion I caused.
 
Ohhh wait. Now that I've looked at this 17 times I think i see where you are going...

I'm really looking at 3/sqrt(5)

Which is 1.34 so, FINAL ANSWER the limit is 1.34!
 
murdazman13 said:
I completely misread the OP. Wow, that's embarrassing. After rereading, i found that 3X2/(5X4)1/2 simplifies to 3/51/2. Does that help any?

Sorry for any confusion I caused.
Click to expand...

No worries! I saw that you misread it originally. No confusion!
 
hopelynnwelch said:
Ohhh wait. Now that I've looked at this 17 times I think i see where you are going...

I'm really looking at 3/sqrt(5)

Which is 1.34 so, FINAL ANSWER the limit is 1.34!
Click to expand...
Your answer is an approximation. The exact answer is 3/sqrt(5). Now can you rationalize this number (that means write the answer w/o any sqrts in the denominator)?
 
Jomo said:
Your answer is an approximation. The exact answer is 3/sqrt(5). Now can you rationalize this number (that means write the answer w/o any sqrts in the denominator)?
Click to expand...


Okay, yes! Just multiply by sqrt(5)/sqrt(5) and get (3*sqrt(5))/5

Thanks Jomo for always being so helpful!
 
hopelynnwelch said:
I need to find the limit as X goes to infinity of (3x^2-1)/(sqrt(5x^4+6))

What I did was make a table and sub in values of 10, 100, 1,000, and 10,000 and I see that as x goes to infinity the function goes to 1.34.

I was curious if I could manipulate the algebraically to come to a conclusion as well but I am not sure.

I was thinking maybe multiply the entire function to get rid on the root in the denominator but I don't know if that really helps at all... Any help appreciated.
Click to expand...
As indicated by others above, the answer to your question is yes, you can manipulate the equation. Take the highest exponent of x in the numerator, in this case 2, extract x2 from the numerator
3 x2 - 1 = x2 (3 - 1/x2)
and note that, as x goes to infinity, the second factor goes to 3. Now do the same thing for the denominator
\(\displaystyle \sqrt{5 x^4 + 6} = \sqrt{ x^4 (5 + 6/x^2)} = x^2 \sqrt{5 + 6/x^2}\)
and the second factor goes to \(\displaystyle \sqrt5\). So we have
\(\displaystyle \frac{3 x^2 - 1}{\sqrt{5 x^4 + 6}} \) ~ \(\displaystyle \frac{x^2}{x^2} \frac{3}{\sqrt5}\)
Since the powers of x are the same in the numerator and denominator we have
\(\displaystyle \lim_{x \to \infty} \frac{3 x^2 - 1}{\sqrt{5 x^4 + 6}} = \frac{3}{\sqrt5} = \frac{3 \sqrt5}{5}\) ~ 1.341641
 
hopelynnwelch said:
I need to find the limit as X goes to infinity of (3x^2-1)/(sqrt(5x^4+6))

What I did was make a table and sub in values of 10, 100, 1,000, and 10,000 and I see that as x goes to infinity the function goes to 1.34.

I was curious if I could manipulate the algebraically to come to a conclusion as well but I am not sure.

I was thinking maybe multiply the entire function to get rid on the root in the denominator but I don't know if that really helps at all... Any help appreciated.
Click to expand...


As x tends to infinity, we can say 1 is negligible compared to 3x^2 and 6 is negligible compared to 5x^4

So it becomes (3x^2)/[sqrt(5x^4)] = (3x^2)/[sqrt(5)x^2] = 3/sqrt(5) ...
 
hopelynnwelch said:
I need to find the limit as X goes to infinity of (3x^2-1)/(sqrt(5x^4+6))

What I did was make a table and sub in values of 10, 100, 1,000, and 10,000 and I see that as x goes to infinity the function goes to 1.34.

I was curious if I could manipulate the algebraically to come to a conclusion as well but I am not sure.

I was thinking maybe multiply the entire function to get rid on the root in the denominator but I don't know if that really helps at all... Any help appreciated.
Click to expand...

\(\displaystyle \displaystyle{\lim_{x\to \infty}\frac{3x^2-1}{\sqrt{5x^4+6}}}\)

\(\displaystyle = \ \displaystyle{\lim_{x\to \infty}\frac{x^2(3 - \frac{1}{x^2})}{x^2\sqrt{5 + \frac{6}{x^4}}}}\)

\(\displaystyle = \ \displaystyle{\lim_{x\to \infty}\frac{3 - \frac{1}{x^2}}{\sqrt{5 + \frac{6}{x^4}}}}\)

\(\displaystyle = \ \displaystyle{\lim_{x\to \infty}\frac{3}{\sqrt{5}}}\)

\(\displaystyle = \ \displaystyle{\frac{3}{\sqrt{5}}}\)
 
One can also argue "For very large x, \(\displaystyle 3x^2\) will be very much larger than -1 so \(\displaystyle 3x^2- 1\) will be very very close to \(\displaystyle 3x^2\). Further, for very large x, \(\displaystyle 5x^4\) wil be much larger than 6 so \(\displaystyle 5x^4+ 6\) will be very very close to \(\displaystyle 5x^4\).

That is, \(\displaystyle \frac{3x^2- 1}{\sqrt{5x^4+ 6}}\) will be very very close to \(\displaystyle \frac{3x^2}{\sqrt{5x^4}}= \frac{3x^2}{\sqrt{5} x^2}= \frac{3}{\sqrt{5}}\).

Cancelling "\(\displaystyle x^2\) from numerator and denominator, as Ishuda and Jomo do, is a more rigorous way of doing this.

By the way, \(\displaystyle \frac{3}{\sqrt{5}}\) is NOT equal to "1.34". Close, but not equal.
 
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