This discussion focuses on isolating the radical in the expression $$n(\sqrt{n^2+4})-n^2$$ and rationalizing it for limit evaluation. Participants clarify the correct interpretation of the expression, emphasizing the importance of factoring out common terms, specifically $$n$$. The conversation highlights the transition from an indeterminate form, such as $$\infty - \infty$$, to a determinate form through rationalization, allowing for limit calculation. The final expression derived is $$\frac{4n}{\sqrt{n^2+4}+n}$$, which can be simplified further for limit evaluation.
PREREQUISITESStudents and educators in mathematics, particularly those focusing on algebra and calculus, will benefit from this discussion. It is especially relevant for individuals looking to improve their understanding of limits and radical expressions.
wishmaster said:How would you isolate the radical in this example:
$$n(\sqrt{n^2+4)}-n^2$$
MarkFL said:It is hard to interpret the given expression since the opening bracket is outside of the radical and the closing bracket is inside. Do you simply mean:
$$n\sqrt{n^2+4}-n^2$$ ?
There is no need to enclose a radical with parentheses, as the radical itself serves as a grouping symbol on its own.
wishmaster said:Im sure about the brackets...it was so written in asignement.
MarkFL said:It is a misprint then...let's write it instead in the more standard form:
$$n\sqrt{n^2+4}-n^2$$
I am thinking this comes from:
$$\lim_{n\to\infty}\left(n\sqrt{n^2+4}-n^2 \right)$$
Am I correct?
Given that we were discussing the rationalization of expressions, what you want to do is factor first...what factor is common to both terms in the expression?
wishmaster said:Its n ?
MarkFL said:Correct! So factoring this common factor out, what do you get?
wishmaster said:$$n(\sqrt{n^2+4}-n)$$
MarkFL said:Correct again! (Star) You're on a roll!
Now, if you are going to rationalize the numerator here*, with what do you think you should multiply this expression?
*Recall that $$f(n)=\frac{f(n)}{1}$$.
wishmaster said:With $$n(\sqrt{n^2+4}+n)$$ ??
MarkFL said:You could do that, but it is simpler to multiply by:
$$1=\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}$$
I recommend that you try both and see that you get the same result, but with what you suggest, you have an extra step of reducing the result. :D
MarkFL said:I would write:
$$n\left(\frac{\sqrt{n^2+4}-n}{1}\cdot\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n} \right)$$
What do you get when you carry out the multiplication within the parentheses?
wishmaster said:$$\frac{n^2+4-n^2}{\sqrt{n^2+4}+n}$$
MarkFL said:Good! Now combine terms in the numerator, and don't forget the factor of $n$ that was outside of the parentheses. What do you wind up with?
wishmaster said:$$\frac{4n}{\sqrt{n^2+4}+n}$$
MarkFL said:Excellent! Now, what were you wanting to do with this expression?
Can i continue from that last expression?MarkFL said:If, as I suspect, you are wanting to take the limit of this expression as $n\to\infty$, then you want to divide each term in the numerator and denominator by $n$.
wishmaster said:Can i continue from that last expression?
MarkFL said:Yes, that is the expression that I meant when dividing each term by $n$.
wishmaster said:I still dot get it what's the most important thing to do with radicals...how to "fight" with them...
I mean,as a question of limits,whats the most important thing to do...?
MarkFL said:As far as limits go, you want to get a determinate form. Forms such as $\infty-\infty$ or $\dfrac{\infty}{\infty}$ are indeterminate. But when you rationalize the numerator in the case of your problem and then divide each term by $n$ you get a determinate form.
wishmaster said:Silly question...what is determinate form?
MarkFL said:A determinate form is one which you can find the limit by substitution of the limiting value for the variable. If you try this with the original form you gave, you get $\infty-\infty$ and you cannot determine anything from this. However, in the final form (after you divide each term by $n$) you find a real number, because it is now a determinate form.
wishmaster said:So i believe this is a thing i should learn...question is how...
MarkFL said:You are on the path...I have noticed a good improvement in your algebra since you have been here. (Yes)
Just keep at it, and gradually the pieces come together. :D
Thank you for the explanation...MarkFL said:Consider the following limit:
$$\lim_{x\to 9}\frac{x-9}{\sqrt{x}-3}$$
If we try to substitute $9$ for $x$, we get $$\frac{0}{0}$$, which is an indeterminate form, as it tells us nothing about the actual value of the limit...it could be anything as far as we know at this point. However, if we factor:
$$\lim_{x\to 9}\frac{\left(\sqrt{x}+3 \right)\left(\sqrt{x}-3 \right)}{\sqrt{x}-3}$$
And now reduce:
$$\lim_{x\to 9}\left(\sqrt{x}+3 \right)$$
Now when we make the substitution, we get a real number:
$$\lim_{x\to 9}\left(\sqrt{x}+3 \right)=\sqrt{9}+3=6$$
This is because we transformed the expression into a determinate form, that is, we can determine the value of the limit by a simple substitution.