What Strategies Can Help Aljosha Tackle Challenging Limits in Math?

[aljosha]

Hello world! My name is Aljosha and I'm in 3rd grade, high school in Serbia.
It's my first post here as you can see. I hope I've chosen the right sub-forum to post this little problem of mine.
I had a test today and these three (below) I couldn't solve.
First one I tried solving by rationalizing (I'm not sure if it's the right term in English) but couldn't finish it as it always appeared to be $$\frac{\infty}{\infty}$$ which is not the solution.
Second one I guess is solved by using property $$\bigg (1+\frac{1}{x} \bigg )^{x} = e$$ but I couldn't solve it. It's probably easy, eh ?

Ahh the third one... If I could only know trig. identity; I think it could be solved similarly as the first one by rationalizing.

Well, my question for you l'equipe is: How would you do it ? I don't need it to be solved, just give me some hints ;)

Thank you in advance!

Here are the limits:

$$\displaystyle\lim_{n \to +\infty} <br /> \frac<br /> {<br /> \sqrt{4n^3+1} + \sqrt{n}<br /> }<br /> {<br /> \sqrt[3]{n^6+2n^2+1} + n<br /> }<br /> }<br /> \\$$


$$\displaystyle\lim_{n \to +\infty}<br /> \bigg (<br /> \frac<br /> {<br /> n-1<br /> }<br /> {<br /> n+5<br /> }<br /> \bigg )^{2n+1}<br /> \\$$


$$\displaystyle\lim_{x \to 0}<br /> \frac<br /> {<br /> \sqrt{\cos{x}} - 1<br /> }<br /> {<br /> x\sin{2x} + x^2<br /> }$$

 

[Lame One]

Your English is fine, and in fact better than most people on the internet these days.

For the first problem (I'm going to do it in 3 posts if you don't mind), try substituting infinity in for n, then just dealing with the infinities with the highest exponent. If you recall, you only need to deal with the infinity which is raised to the highest power. After a bit of work, you should be able to figure that out. I'll have the next one up for you shortly.

EDIT: Sorry, but I will not be able to get the other two up until later today. I am at a school computer, which doesn't recognize the advanced toolbox. Therefore, I cannot write easy to read mathematic text. Hopefully, I will be able to help you soon, or, even better, someone else can jump in. Sorry!

 

[Mark44]

Your English is fine, and in fact better than most people on the internet these days.

I agree completely.

For the first problem (I'm going to do it in 3 posts if you don't mind), try substituting infinity in for n, then just dealing with the infinities with the highest exponent.

This is very bad advice. You should never substitute infinity and try to do arithmetic.

Instead, factor n³ out of each term in both radicals in the numerator, and then bring it out of the radical so that you have n^3/2 outside.
In the denominator, factor n⁶ out of each term in the first radical, as well as out of the second term. In the numerator you should get n^3/2 times some other stuff whose limit can be evaluated easily. In the denominator you should get n² times some stuff that can also be evaluated easily.

If you recall, you only need to deal with the infinity which is raised to the highest power. After a bit of work, you should be able to figure that out. I'll have the next one up for you shortly.

EDIT: Sorry, but I will not be able to get the other two up until later today. I am at a school computer, which doesn't recognize the advanced toolbox. Therefore, I cannot write easy to read mathematic text. Hopefully, I will be able to help you soon, or, even better, someone else can jump in. Sorry!

 

[Lame One]

I did not suggest arithmetic. By substituting infinity, it allows for you to reasonably eliminate most constants from the equation. Without doing this, you would be left with finding the square root of a polynomial, which is, of course, more difficult. Substituting in infinity and only dealing with the one with the highest exponent let's you deal with a single term, which is nearly always much simpler.

 

[Mark44]

I did not suggest arithmetic. By substituting infinity, it allows for you to reasonably eliminate most constants from the equation.

This is not how limits at infinity are dealt with. What you said was "...dealing with the infinities with the highest exponent..."
Here it seems that you are implying that, for example, $\infty^3$ is somehow larger than $\infty^2$.

Without doing this, you would be left with finding the square root of a polynomial, which is, of course, more difficult.

Perhaps more difficult, but not much, and in any case, valid to do.

Here's an example that is similar to the OP's problem.
$$\lim_{x \to \infty}\frac{\sqrt{x^3 + 2x}}{4x^2 + 3x}$$
$$=\lim_{x \to \infty}\frac{x^{3/2}\sqrt{1 + 2/x^3}}{x^2(4 + 3/x}$$
$$=\lim_{x \to \infty}\frac{\sqrt{1 + 2/x^3}}{x^{1/2}(4 + 3/x} = 0$$

For any finite x, x^(3/2)/x^2 = 1/x^(1/2), which approaches 0 as x approaches infinity. The radical in the numerator approaches 1 as x approaches infinity, and the expression 4 + 3/x in the denominator approaches 4.

Substituting in infinity and only dealing with the one with the highest exponent let's you deal with a single term, which is nearly always much simpler.

 

[Lame One]

In a way, $\infty^3$ is larger than $\infty^2$. In the case of finding the limit of $\frac{100x^2}{20x^3}$, for instance. I realize that you can easily eliminate the x's here, but that's not the point. Either way, this is getting off topic. I don't want to end up being the source of an argument that turns this into a spam thread.

 

[aljosha]

Your English is fine, and in fact better than most people on the internet these days.

I agree completely.

Thank you guys on kind words.

I made some progress, I've solved first one as @Mark44 suggested and second one using property I've wrote above - it took some algebraic fiddling but hey, I did it!
Now I'm left with the third one, any suggestions how to solve it ?
P.S

EDIT: Sorry, but I will not be able to get the other two up until later today. I am at a school computer, which doesn't recognize the advanced toolbox. Therefore, I cannot write easy to read mathematic text. Hopefully, I will be able to help you soon, or, even better, someone else can jump in. Sorry!

I'm not in a hurry just so you know ;)

 

[Dick]

Thank you guys on kind words.

I made some progress, I've solved first one as @Mark44 suggested and second one using property I've wrote above - it took some algebraic fiddling but hey, I did it!
Now I'm left with the third one, any suggestions how to solve it ?
P.S

I'm not in a hurry just so you know ;)

I would start by multiplying numerator and denominator by sqrt(cos(x))+1. Then if you know the series expansions of sin(x) and cos(x) around x=0, it's pretty easy. Otherwise, mess with it and use trig limits you do know, like sin(x)/x ->1 as x ->0.

 

[aljosha]

I decided to upload texes because somebody might need reference in the future. And big thanks to all of you who posted here, it really helped ;) I love this forum ^_^

The second one:

$$\lim_{n \to \infty}<br /> \big{<br /> ({{n-1} \over {n+5}})^{2n+1}<br /> }$$
$$\text{\emph{we can write that as }} <br /> \displaystyle\lim_{n \to \infty}<br /> \big{({{n+5-6} \over {n+5}})^{2n+1}}$$
$$\text{\emph{ after which we have }}<br /> \displaystyle\lim_{n \to \infty}<br /> \big{(1-{{6} \over {n+5}})^{2n+1}}$$
$$\text{\emph{ work towards using this property }}<br /> \displaystyle\lim_{n \to \infty}<br /> \big{<br /> {(1+{{1} \over {x}})^{x}}=e<br /> }$$
$$\text{\emph{ now we can write our expression }}<br /> \<br /> \displaystyle\lim_{n \to \infty}<br /> \big{<br /> {(1+{{1}\over{{n+5}\over6}}})^{ ({{6}\over{n+5}})({{n+5} \over{6}}) ({2n+1})}<br /> }$$
$$\text{\emph{ which is: }}<br /> \\<br /> e^{\displaystyle\lim_{n \to \infty}\bigg{<br /> \frac{n(12 + \frac{6}{n})} {n(1+\frac{5}{n})}<br /> }}<br /> =e^{-12}$$

Third one:

$$\displaystyle\lim_{x \to 0}<br /> \frac{\sqrt{\cos{x}} - 1}{x\sin{2x} + x^2}$$

$$\displaystyle\lim_{x \to 0}<br /> \frac{\sqrt{\cos{x}} - 1}{ x\sin{2x} + x^2} \cdot \bigg / \frac{\sqrt{\cos{x}}+1}{\sqrt{\cos{x}}+1}$$

$$\text{ }<br /> \displaystyle\lim_{x \to 0}<br /> \big{\cos{x} - 1 \over \sin{2x} \sqrt{\cos{x}} + x \sin{2x} + x^2 \sqrt{\cos{x}} + x^2 }$$

$$\text{\emph{Using trigonometric identity for}}<br /> \cos{x}-1 = -2\sin{x \over 2}\sin{x \over 2}$$

$$\text<br /> {<br /> \emph{and by using limits property}<br /> }<br /> \displaystyle\lim_{x \to 0} <br /> \big{{\frac{\sin{x}}{x} = 1}}$$

$$\text{\emph{ we get }}<br /> \displaystyle\lim_{x \to 0}<br /> \big{<br /> {-1x^2 \over 2} \over x^2(2\sqrt{\cos{x}}+2+\sqrt{\cos{x}} + 1)<br /> }$$

$$\text{\emph{ by canceling }} x^2 <br /> \displaystyle\lim_{x \to 0}<br /> \big{<br /> {x^2({-1\over 2})} \over {x^2(2\sqrt{\cos{x}}+2+\sqrt{\cos{x}}+1)}<br /> }$$

$$\text{\emph{ we get the solution }}<br /> \displaystyle\lim_{x \to 0}<br /> \big{ {{-1 \over 2} \over 6} = {-1 \over 12} }$$