How can I prove this limit? \lim \limits_{x \to \infty} \sqrt{x^2-4x+5}

hestyllvql

hestyllvql

Answered question

2022-04-22

How can I prove this limit?
limxx24x+5=x2

Answer & Explanation

Aliana Sexton

Aliana Sexton

Beginner2022-04-23Added 20 answers

Step 1
Let me start with what you did wrong: This expression
limxx24x+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 ). 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
=x2
It is only true if x=, which is not a value that normally x is allowed to even take. (That is why we talk about the "limit as x'' and not the "value at x='') So your question as stated didn't make sense.
What should you have written? You actually meant that the two functions
f(x)=x24x+5 and g(x) are asymptotic as x There are different ways to define this. The comments have it interpreted as
limx[x24x+5(x2)]=0
This is somewhat tricky to show. herb steinberg almost has it, but doesn't justify the claim that it is x2+O(1x)
What you should start with is a change of variable: let t=x2, noting that t as x and vice versa. Then it becomes
limtt2+1t
=limt1+1t211t
=lims0+1+s21s=f(0)
where f(s)=1+s2 It is easily checked that f(0)=0
A somewhat looser definition is
limx[x24x+5x2]=1
Which is easy to prove, since for x>2
x24x+5x2=1+1(x2)2

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