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What is #lim_(x->oo) (x^2-x)# ?

Answer 1

Evelyn Agard

#lim_(x->oo) (x^2-x) = oo#

If we are a little too incautious and treat #oo# as if it's a number, then we might try:

#lim_(x->oo) (x^2-x) = oo*oo - oo = oo - oo = ?#

Instead, note that:

#x^2-x = x(x-1)#

and both:

#lim_(x->oo) x = oo#

#lim_(x->oo) x-1 = oo#

So we find:

#lim_(x->oo) (x^2-x) = lim_(x->oo) (x(x-1))#

#= lim_(x->oo) x * lim_(x->oo) (x-1) = oo*oo = oo#

Alternatively, let's try a slightly more formal approach...

Given any #N > 0#

then for any #x > N+2# we have:

#x-1 > N+2-1 = N+1 > 1#

and

#x > N+2-1 = N+1 > N#

Hence:

#x^2-x = x(x-1) > N * 1 = N#

So we have shown that for any #N > 0# there is some #M > 0# (e.g. #M = N+2#) such that for all #x > M#, #(x^2-x) > N#.

Therefore:

#lim_(x->oo) (x^2-x) = oo#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is #lim_(x-&gt;oo) (x^2-x)# ?

What is #lim_(x->oo) (x^2-x)# ?