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We have the function $$\dfrac{\sqrt{n^4 + 100}}{4n}$$

I think the best method is by dividing by $n$, but I have no idea what that yields, mainly because of the square root.

LimitsMate
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5 Answers5

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Dividing by $n$ is a great idea: Note that $$\sqrt{n^4 + 100}\cdot\frac 1n = \sqrt{(n^4 + 100)\cdot\frac 1{n^2}} = \sqrt{n^2 + \frac {100}{n^2}}$$

So your limit is, after dividing numerator and denominator by $n$:

$$\lim_{n\to \infty} \frac{\sqrt{n^2 + \frac {100}{n^2}}}{4} \to \infty$$

amWhy
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I assume you mean the limit as $n\to \infty$. In that case, the limit is easily seen to be infinity as follows:

The numerator scales as $\sim n^2$ for large $n$, but the denominator scales as $n$ only.

The ratio thus scales as $\sim n$.

So the limit as $n\to \infty$ will be $\infty$.

Gandhi Viswanathan
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$\text{Note that $\sqrt{n^4+100} > \sqrt{n^4}$. Hence, we have}$ $$\dfrac{\sqrt{n^4+100}}{4n} > \dfrac{\sqrt{n^4}}{4n} = \dfrac{n^2}{4n} = \dfrac{n}4$$ $\text{I trust you can finish it from here.}$

  • So is it always advisable to get rid of the numbers in such cases (since that don't contribute anything)? – LimitsMate May 14 '13 at 19:17
  • @LimitsMate Yes. Most often is is good to throw of the "lower" order terms and you can conclude from the rest. –  May 14 '13 at 19:18
  • Perhaps I have found an example where your method doesn't work? $\dfrac{\sqrt{100+4^n}}{10+2^n}$, if you take $\dfrac{\sqrt{4^n}}{10+2^n}$ you get the incorrect limit of 1. Am I correct? – LimitsMate May 14 '13 at 19:43
  • @LimitsMate, it's not so much a matter of "throwing away" the constant factor with disregard. This can be done here because clearly $\sqrt{n^4+100}>\sqrt{n^4}$ – Ray May 14 '13 at 19:54
  • @Ray so? In my example it is also true but it still doesn't work – LimitsMate May 14 '13 at 20:00
  • @LimitsMate The limit in the example you quote is $\color{red}{\text{indeed $1$}}$. –  May 14 '13 at 20:03
  • @user17762 Wait what? Wolfram alpha told me the first limit is 0 (and I thought so too) and the second is obviously 1. – LimitsMate May 14 '13 at 20:06
  • @user17762 Never mind, your method is all powerful sir! – LimitsMate May 14 '13 at 20:19
  • @LimitsMate $\sqrt{4^n} = 2^n$ and $\lim_{n \to \infty} \dfrac{2^n}{2^n+10} = 1$ And $$\lim_{n \to \infty} \dfrac{\sqrt{4^n+100}}{2^n+10} = \lim_{n \to \infty} \dfrac{2^n\sqrt{1+\dfrac{100}{4^n}}}{2^n\left(1+\dfrac{10}{2^n}\right)} = \lim_{n \to \infty} \dfrac{\sqrt{1+\dfrac{100}{4^n}}}{\left(1+\dfrac{10}{2^n}\right)} = \dfrac{\lim_{n \to \infty} \sqrt{1+\dfrac{100}{4^n}}}{\lim_{n \to \infty} \left(1+\dfrac{10}{2^n}\right)} = \dfrac{\sqrt{1+\lim_{n \to \infty} \dfrac{100}{4^n}}}{ \left(1+\lim_{n \to \infty}\dfrac{10}{2^n}\right)}$$ –  May 14 '13 at 20:20
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You have that $$\frac{\sqrt{n^4 + 100}}{4n} = \frac{n^2\sqrt{1 + \frac{100}{n^4}}}{4n}=\frac{n\sqrt{1 + \frac{100}{n^4}}}{4}$$ and $\frac{\sqrt{1 + \frac{100}{n^4}}}{4}\xrightarrow[n\to\infty]{}\frac{1}{4}$.

Clement C.
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  • Sorry Clement, that answer is not right, the limit is infinite as stated by Gandhi. The product of the limits is the limit of the products is only valid if both limits exist. In your case, the radical limit does not exist. – imranfat May 14 '13 at 18:56
  • @imranfat: I'm sorry to say that you don't seem to have read what I wrote. The overall limit is $+\infty$, but what I did write is a way to prove it; more precisely, to prove that the quantity is equivalent to $\frac{n}{4}$. – Clement C. May 14 '13 at 18:58
  • I must have read it too fast...but I do wonder if the quantity is equal to n/4 then what will happen with this quantity if n goes to infinity. – imranfat May 15 '13 at 15:22
  • What it says is basically that it will go to infinity, but at a rate comparable to $n/4$ (the two quantities are equivalent). That is, if you plot the points on a graph, then the points of the sequence will be distributed eventually arbitrarily close to the line $y=x/4$. – Clement C. May 15 '13 at 15:29
  • Yes, that makes sense. I subtracted an+b from the given limit, multiplied this difference with its conjugate, made it into one fraction with 16n^4 as its common Denom and lo and behold: the coefficient of n^4 in the Num is 1-16n^2 which will have to be zero for the limit to exist and so a = 1/4 pops right out!! – imranfat May 15 '13 at 16:34
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Perhaps another method I overlooked, stupidly enough:

$$ \dfrac{\sqrt{n^4 + 100}}{4n} = \dfrac{n^4 + 100}{4n^5 + 400n} = \dfrac{ n^4 + 100}{4n \cdot n^4 + 400n} = \dfrac{101}{404n} = \dfrac{1}{4n}$$

I'm not sure it is correct, I played around a bit with the algebra.

LimitsMate
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  • A very easy way to verify is to put both expressions as functions in the graphing calculator. It turns out that the algebra is not correct because the graphs do not match – imranfat May 15 '13 at 15:24