Is it allowed to evaluate a limit this way ? And if not, why ?

[ronfya]

Hello All,

Are both of the following solutions to evaluate limits acceptable ? Or is there one that is "not allowed" for some reason ?
Thanks

Using product conjugate pattern




or in a simpler way


(1): because x^2+constant ~ x^2 when x --> inf
(2): because sqrt(x^2) = x when x>0

 

[Dr.Peterson]

Hello All,

Are both of the following solutions to evaluate limits acceptable ? Or is there one that is "not allowed" for some reason ?
Thanks

Using product conjugate pattern

View attachment 36548


or in a simpler way

View attachment 36549
(1): because x^2+constant ~ x^2 when x --> inf
(2): because sqrt(x^2) = x when x>0

What you write in the second method assumes that you can replace an expression within a limit with another expression that has the same limit.

Do you have any theorem that says you can do that? If not, then you can't do it.

 

[blamocur]

Hello All,

Are both of the following solutions to evaluate limits acceptable ? Or is there one that is "not allowed" for some reason ?
Thanks

Using product conjugate pattern

View attachment 36548


or in a simpler way

View attachment 36549
(1): because x^2+constant ~ x^2 when x --> inf
(2): because sqrt(x^2) = x when x>0

You second method is not valid. Using the same logic I could write
[math]\lim_{x\rightarrow\infty} ((x-2) - (x+1)) = \lim_{x\rightarrow\infty} (x - x) = 0[/math]But this is just an illustration. The actual reason is in @Dr.Peterson's post: you need to show/prove that your transformation is a valid one.

 

[Qwertyuiop[]]

You second method is not valid. Using the same logic I could write
[math]\lim_{x\rightarrow\infty} ((x-2) - (x+1)) = \lim_{x\rightarrow\infty} (x - x) = 0[/math]But this is just an illustration. The actual reason is in @Dr.Peterson's post: you need to show/prove that your transformation is a valid one.

But the conjugate method is correct right? Because we are just multiplying by 1.

 

[blamocur]

But the conjugate method is correct right? Because we are just multiplying by 1.

I agree.

 

[Dr.Peterson]

But the conjugate method is correct right? Because we are just multiplying by 1.

Yes, and in particular, every step is supported by a theorem.

Most steps are just rewriting the expression (not changing the function itself, and therefore not changing the limit); at the end, it is not technically valid to write that you are dividing by infinity, but there is a theorem that if a function can be written as a quotient of functions whose limits have the form "finite number / infinity", then the limit is 0.