# Limit of the sequence $\lim_{n\rightarrow \infty}n\left ( 1-\sqrt{1-\frac{5}{n}} \right )$, strange result

$\lim_{n\rightarrow \infty}n\left ( 1-\sqrt{1-\frac{5}{n}} \right )$

$\lim_{n\rightarrow \infty} n *\lim_{n\rightarrow \infty}\left ( 1-\sqrt{1-\frac{5}{n}} \right ) = \infty * \left ( 1-\sqrt{1-0} \right ) = \infty * 0 = 0$

Did I do it correctly?

Problem is when I use my calculator and put big values for $n$, I get $2,5$ as result (if I take very very big values, it's $0$ :D).

Anyway, this made me feel a bit insecure and that's why I'm asking if I did it right.

In short: no, you did not do it correctly. The reason is that $\infty\cdot 0$ is one of the indeterminate forms, you cannot conclude it's equal to zero.

The rationale being: it can be anything. For instance, \begin{align} \underbrace{n}_{\to \infty}\cdot \underbrace{\frac{1}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} 1\\ \underbrace{n^2}_{\to \infty}\cdot \underbrace{\frac{1}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} \infty\\ \underbrace{n}_{\to \infty}\cdot \underbrace{\frac{1}{n^2}}_{\to 0} &\xrightarrow[n\to\infty]{} 0\\ \underbrace{n}_{\to \infty}\cdot \underbrace{\frac{\cos n}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} \text{ nothing (no limit)} \end{align}

To solve your indeterminate form, you have to use other techniques to "remove" that issue. Taylor expansions, multiplication by a conjugate, L'Hôpital rule... there are known and systematic methods to do so.

E.g., with Taylor expansions: when $u\to 0$, $$\sqrt{1-5u} = 1-\frac{5}{2}u + o(u)$$ so applying it with $u\stackrel{\rm def}{=}\frac{1}{n}\xrightarrow[n\to\infty]{}0$, \begin{align} n\left(1- \sqrt{1-\frac{5}{n}}\right) &= \frac{1}{u}\left(1- \sqrt{1-5u}\right) = \frac{1}{u}\left(1- 1 + \frac{5}{2}u + o(u)\right) = \frac{1}{u}\left(\frac{5}{2}u + o(u)\right)\\ &= \frac{5}{2} + o(1) \xrightarrow[n\to\infty]{}\frac{5}{2}. \end{align}

Addendum: why does your calculator do that? The correct answer is indeed $\frac{5}{2}=2.5$. Yet, for "very big $n$", your calculator will return $0$ instead... and that boils down to machine precision. I assume it starts by computing the term $\sqrt{1-\frac{5}{n}}$, then plugs in the result to get $1-\sqrt{1-\frac{5}{n}}$, and finally multiplies what remains by $n$ and outputs the value. But for very large $n$, due to machine precision and rounding errors, the first step will be so close to $1$ that the calculator will just compute $1$ instead of $\sqrt{1-\frac{5}{n}}\approx 1 - \frac{5}{2n}$. So then, $1-1=0$, and no matter what the last step should do, it returns $n\cdot 0=0$.

• Very well explained. +1 – Mark Viola Jul 22 '16 at 16:38
• Wow thanks a lot for your answer man! Taylor polynoms, I heard of these but nothing else so far. Then I should learn them to understand your post completely and to solve tasks like that. I have just tried to solve it with L'Hôpital (good to know you can use it for multiplications as well, thought it was only made for fractions). But I always end up with $0 * \infty$...Never mind that. And thank you very much for the clarification with the calculator ^.^ – cnmesr Jul 22 '16 at 16:52
• Glad i could help! If you feel more comfortable with this, you can also see your limit as computing a derivative. Set $$f(x) = \sqrt{1-5x}$$ for $x\in[-1,1]$ and look at $\frac{f(x)-f(0)}{x-0} = \frac{\sqrt{1-5x}-1}{x}$. The limit at $0$ will be $f'(0)$... but does the expression ring a bell (e.g., considering $x=\frac{1}{n}$)? – Clement C. Jul 22 '16 at 16:55

$$n-n\sqrt{1-\frac{5}{n}}=\frac{(n-n\sqrt{1-\frac{5}{n}})(n+n\sqrt{1-\frac{5}{n}})}{n+n\sqrt{1-\frac{5}{n}}}=\frac{5}{1+\sqrt{1-\frac{5}{n}}}\rightarrow \frac{5}{2}, \space \text{as} \space \space n \rightarrow \infty.$$

• Damn it I haven't thought about that, using the binomial formula! >.< Thanks so much for this! – cnmesr Jul 22 '16 at 18:30
• @cnmesr This is not really an application of the Binomial formula, rather an instance of "rationalizing the numerator." – Clement C. Jul 22 '16 at 18:48
• But what he used is the 3th binomial formula, isn't it? $(a+b)*(a-b)$ Ok I think we may call basic idea is binomial formula, not sure.. :D But now I understood both of your answers and I'm so happy I did, now I can relax and continue learning tomorrow. Thanks a lot again to both of you! :) – cnmesr Jul 22 '16 at 18:51

No, you can't do it like that: when you have a product where one factor has limit $\infty$ and the other has limit $0$, you cannot apply a theorem on products of limits.

The theorem helps when both limits are finite and you just multiply them; it also works when one limit is $\infty$ (or $-\infty$) and the other one is either infinite or *finite and not $0$”. Also in this case you can “multiply”: the limit will be either $\infty$ or $-\infty$, depending on the signs of the factors.

Instead of the sequence (which is implied by the use of $n$), try and find the limit of the function: $$\lim_{x\to\infty}x\left(1-\sqrt{1-\frac{5}{x}}\right)= \lim_{t\to0^+}\frac{1-\sqrt{1-5t}}{t}$$ with the substitution $x=1/t$. This limit is much easier to manage; if it exists, then the sequence will have the same limit. Note however that the limit of the function may not exist whereas the limit of the sequence exists. Not in this case: take your pick below.

## 1. Rationalization

$$\lim_{t\to0^+}\frac{1-\sqrt{1-5t}}{t} = \lim_{t\to0^+}\frac{1-(1-5t)}{t(1+\sqrt{1-5t})}= \lim_{t\to0^+}\frac{5)}{1+\sqrt{1-5t}}=\frac{5}{2}$$

## 2. Taylor expansion

$$\lim_{t\to0^+}\frac{1-\sqrt{1-5t}}{t} = \lim_{t\to0^+}\frac{1-(1-\frac{5}{2}t+o(t))}{t} = \lim_{t\to0^+}\left(\frac{5}{2}+o(1)\right)=\frac{5}{2}$$

## 3. Derivative

The limit is the derivative at $0$ of $f(t)=1-\sqrt{1-5t}$ and $$f'(t)=\frac{5}{2\sqrt{1-5t}}$$ so $$f'(0)=\frac{5}{2}$$