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question on limits of a real function as x -> infinity

hello there,
I am looking at the following question:

" Let f be a real valued function such that f(x) = x^4 - 3x^2. Show that f(x) tends to infinity as x tends to infinity."

So, by definition, we know that f(x)-> infinity as x-> infinity if

" For all A > 0, there exists K E R such that f(x) > A whenever x > K ."

Now for x > 2 , x^2 > 4 => f(x) = x^2(x^2 - 3) > x^2.

Hence , f(x) > a provided x > max{ sqrt A, 2}. Takin K as max{ sqrt A, 2} gives f(x) > A whenever x > K .

I am having trouble understanding this. I can see where the value of K = 2 comes from, but how do you come to K being the sqrt A ??? And reach the conclusion that K = max {sqrt A, 2}??

Also, is there a specific way to go about doing this type of question ?

Thank you very much for your time !

Reply 1

yusufu

what level is this for?

Reply 2

KAISER_MOLE

looking at his previous posts, I'm gonna guess he is an undergraduate sort of guy.

....oh crumbs, MotD has started! *scurries downstairs

Reply 3

dvs

You want f(x) > A, that is:
x^2(x^2 - 3) > A

For x>2, we have x^2 - 3 > 1, and hence:
f(x) = x^2(x^2 - 3) > x^2

So if f(x) > A, then x^2 > A will give us that. Thus, we have that whenever x > sqrt(A) and x > 2, then:
f(x) > A
for any A in R.

So if x > max{sqrt(A), 2}, then both x > sqrt(A) and x > 2 are satisfied, and hence f(x) > A.

Reply 4

Jonny W

f(x) = x^2 (x^2 - 3)

will be >= A if:
- the red thing is >= A, and
- the blue thing is >= 1.

The first statement is true if x >= sqrt(A).
The second statement is true if x >= 2.
Both statements are true if x >= max[sqrt(A), 2].