hestyllvql

2022-04-22

How can I prove this limit?

${\mathrm{lim}}_{x\to \mathrm{\infty}}\sqrt{{x}^{2}-4x+5}=x-2$

Aliana Sexton

Beginner2022-04-23Added 20 answers

Step 1

Let me start with what you did wrong: This expression

$\underset{x\to \mathrm{\infty}}{lim}\sqrt{{x}^{2}-4x+5}$

does not depend on a variable x. There is no x in it. The x you see is a dummy variable. That x exists only within that notation, and is there only to make the notation work. In computing, they call that "scope". x comes into existence with the "lim" and goes back out of existence after the 5 and the closing of the square root.

The expression represents a single value (though in this case, not a member of the real numbers, but$\mathrm{\infty}$ ). It is not a variable expression. It is not a function. It is a single value. On the other hand, "x-2" is a variable expression. It takes on different values for different values of x, which is a free variable here. It has no relation to the dummy variable x that appears on the left and does not exist on the right.

So what you wrote reduces to

$\mathrm{\infty}=x-2$

It is only true if$x=\mathrm{\infty}$ , which is not a value that normally x is allowed to even take. (That is why we talk about the "limit as $x\to \mathrm{\infty}$ '' and not the "value at $x=\mathrm{\infty}$ '') So your question as stated didn't make sense.

What should you have written? You actually meant that the two functions

$f\left(x\right)=\sqrt{{x}^{2}-4x+5}$ and g(x) are asymptotic as $x\to \mathrm{\infty}$ There are different ways to define this. The comments have it interpreted as

$\underset{x\to \mathrm{\infty}}{lim}[\sqrt{{x}^{2}-4x+5}-(x-2)]=0$

This is somewhat tricky to show. herb steinberg almost has it, but doesn't justify the claim that it is$x-2+O\left(\frac{1}{x}\right)$

What you should start with is a change of variable: let$t=x-2$ , noting that $t\to \mathrm{\infty}$ as $x\to \mathrm{\infty}$ and vice versa. Then it becomes

$\underset{t\to \mathrm{\infty}}{lim}\sqrt{{t}^{2}+1}-t$

$=\underset{t\to \mathrm{\infty}}{lim}\frac{\sqrt{1+\frac{1}{{t}^{2}}}-1}{\frac{1}{t}}$

$=\underset{s\to 0+}{lim}\frac{\sqrt{1+{s}^{2}}-1}{s}={f}^{\prime}\left(0\right)$

where$f\left(s\right)=\sqrt{1+{s}^{2}}$ It is easily checked that ${f}^{\prime}\left(0\right)=0$

A somewhat looser definition is

$\underset{x\to \mathrm{\infty}}{lim}\left[\frac{\sqrt{{x}^{2}-4x+5}}{x-2}\right]=1$

Which is easy to prove, since for$x>2$

$\frac{\sqrt{{x}^{2}-4x+5}}{x-2}=\sqrt{1+\frac{1}{{(x-2)}^{2}}}$

Let me start with what you did wrong: This expression

does not depend on a variable x. There is no x in it. The x you see is a dummy variable. That x exists only within that notation, and is there only to make the notation work. In computing, they call that "scope". x comes into existence with the "lim" and goes back out of existence after the 5 and the closing of the square root.

The expression represents a single value (though in this case, not a member of the real numbers, but

So what you wrote reduces to

It is only true if

What should you have written? You actually meant that the two functions

This is somewhat tricky to show. herb steinberg almost has it, but doesn't justify the claim that it is

What you should start with is a change of variable: let

where

A somewhat looser definition is

Which is easy to prove, since for

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