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

Question

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

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

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.

Answer 1

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.

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

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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 $n\cdot 0=0$.

Answer 2

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

Answer 3

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 $$ f'(0)=\frac{5}{2} $$