Confused about process for finding limits at infinity?

**DannyMath** · December 12th 2010, 01:26 PM

Hi, I'm practicing for Calc test and the solutions to a problem are throwing me off. I know that to find a limit as x approaches infinity you can divide everything by the highest power x in the denominator but the solution steps show the person doing this:

I see that they divided everything by x^2, including everything under the square root. But wouldn't you need to divide the stuff under the square by x^4 since that's sqrt(x^4) is equivalent to x^2? Any help is appreciated

EDIT: Also the original question actually asks for the limit as x approaches NEGATIVE infinity. I'm looking at this problem after posting and I'm frankly lost on the steps, but I will keep working on it.

**Plato** · December 12th 2010, 02:00 PM

I really don't understand what you are asking from us.
The algebra in the solution is straightforward.
So what can we say?

**dwsmith** · December 12th 2010, 02:15 PM

$\displaystyle \lim_{x\to\infty}\frac{x^2*(2x)^3}{x^4*\sqrt{x^2}} \Rightarrow\lim_{x\to\infty}\frac{8x^5}{x^5}=8$

Just take the highest powers from each product in both the denominator and numerator and then simplify.

**adkinsjr** · December 12th 2010, 02:39 PM

edit:

I had an error in my algebra...

**Plato** · December 12th 2010, 02:45 PM

Originally Posted by adkinsjr

The algebra doesn't seem simple to me. I would just use dwsmith's method of highest powers. But I'm not following the algebra in the solution either.

Well that is why it is called Pre-Calculus.
You need basic algebra to do calculus.

**adkinsjr** · December 12th 2010, 02:49 PM

Originally Posted by Plato

Well that is why it is called Pre-Calculus.
You need basic algebra to do calculus.

I've been through calcs 1-3 and I obtained A's in all 3.

**adkinsjr** · December 12th 2010, 03:12 PM

Ok, I see what your problem is now...

Look at it like this:

$\lim_{x->\infty}\frac{x^2(2x+1)^3}{(x^2+2)^2\sqrt{x^2+2x+3 }}=\lim_{x->\infty}\frac{x^2x^3\frac{1}{x^3}(2x+1)^3}{x^5\fra c{1}{x^4}(x+2)^2\frac{1}{x}\sqrt{x^2+2x+3}}$

This seems to be what they did, they didn't divide the square root by x^2, they just divided it by x... However, I still think they messed up after that. The algebra they did just doesn't make sense...not to me anyways...

**Archie Meade** · December 12th 2010, 03:14 PM

Originally Posted by DannyMath

Hi, I'm practicing for Calc test and the solutions to a problem are throwing me off. I know that to find a limit as x approaches infinity you can divide everything by the highest power x in the denominator but the solution steps show the person doing this:

I see that they divided everything by x^2, including everything under the square root. But wouldn't you need to divide the stuff under the square by x^4 since that's sqrt(x^4) is equivalent to x^2? Any help is appreciated

EDIT: Also the original question actually asks for the limit as x approaches NEGATIVE infinity. I'm looking at this problem after posting and I'm frankly lost on the steps, but I will keep working on it.

$\displaystyle\frac{x^2(2x+1)^3}{(x^2+2)^2\sqrt{x^2 +2x+3}}=\frac{x^2\left[x\left(2+\frac{1}{x}\right)\right]^3}{\left[x^2\left(1+\frac{2}{x^2}\right)\right]^2\left[\sqrt{x^2}\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}\right]}$

$=\displaystyle\frac{x^2x^3\left(2+\frac{1}{x}\righ t)^3}{x^4x\left(1+\frac{2}{x^2}\right)^2\sqrt{1+\f rac{2}{x}+\frac{3}{x^2}}}$

so there is a little error in the text but it is inconsequential to the calculation of the limit.

**adkinsjr** · December 12th 2010, 03:26 PM

Originally Posted by Archie Meade

$\displaystyle\frac{x^2(2x+1)^3}{(x^2+2)^2\sqrt{x^2 +2x+3}}=\frac{x^2\left[x\left(2+\frac{1}{x}\right)\right]^3}{\left[x^2\left(1+\frac{2}{x^2}\right)\right]^2\left[\sqrt{x^2}\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}\right]}$

$=\displaystyle\frac{x^2x^3\left(2+\frac{1}{x}\righ t)^3}{x^4x\left(1+\frac{2}{x^2}\right)^2\sqrt{1+\f rac{2}{x}+\frac{3}{x^2}}}$

so there is a little error in the text but it is inconsequential to the calculation of the limit.

There you go, thankyou.. I see where they messed up now. I couldn't spot that...

**DannyMath** · December 12th 2010, 05:32 PM

Originally Posted by Plato

Well that is why it is called Pre-Calculus.
You need basic algebra to do calculus.

I don't understand what the point is in posting this type of stuff. I doubt that you believe every person who has passed Pre-Cal would find this straightforward.

Originally Posted by Archie Meade

so there is a little error in the text but it is inconsequential to the calculation of the limit.

Thanks for the helpful info.

**DannyMath** · December 13th 2010, 07:47 AM

Hi again all. I'm doing this problem again but this time it's as x goes to negative infinity and I see that the graph shows that as x goes to negative infinity, the limit is -8. Is this because when you are going from sqrt(x^2) to x you need to add a negative sign to keep the expression true? Because that's the only way I found I would get -8 rather than 8. Thanks again

**Plato** · December 13th 2010, 08:33 AM

Look carefully at the original problem.
There are four factors: $x^2,~(2x+3)^3,~(x^2+1)^2~\&~\sqrt{x^2+2x+3}$ .
As $x\to -\infty$ all but one of those factors are positive, one is negative.
So the answer is negative. The algebra part of the workings simply helps us find the numerical value. See post #3. But it does not give the correct sign. That comes from an analysis of the original factors.

**DrSteve** · December 13th 2010, 08:36 AM

$\sqrt{x^2} = -x$ for negative values of x.

In general, $\sqrt{x^2} =|x|$

**Archie Meade** · December 13th 2010, 08:43 AM

Originally Posted by DrSteve

$\sqrt{x^2} = -x$ for negative values of x.

In general, $\sqrt{x^2} =|x|$

Yes, the equation contains the positive square root.
See also post 12 for the alternative evaluation of the sign.

Post 8 takes x as positive only.

**DrSteve** · December 13th 2010, 09:02 AM

Yes - that's true. The equation does contain the positive square root. What I've given is the definition of the positive square root of x (note that -x is a positive number when x<0).

If you're using the shortcut method to compute the limit, then this information is not needed. But if you want to write out the details more rigorously (as was done in the very first post), then this definition will be needed.

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