Calculate limit (x^2+2x+1)/(2x^2-2x-1), x->+infinity

4 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

limit (x^2+2x+1)/(2x^2-2x-1)  when x--> inf

We observe that he highest power for the numerator and denominator is x^2, then we divide both functions by x^2.

==> lim x^2(1+2/x+1/x^2)/lim x^2(2-2/x-1/x^2)

==> lim (1+2/x+1/x^2)/lim (2-2/x-1/x^2) = (1+0+0)/(2-0-0)=1/2

then the limit when x--> inf is 1/2

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

In order to calculate the limit of a rational function, when x tends to +inf., we'll divide both, numerator and denominator, by the highest power of x, which in this case is x^2.

We'll have:

lim[(x^2+2x+1)/(2x^2-2x-1)]= lim(x^2+2x+1)/lim(2x^2-2x-1)

lim x^2(1 + 2/x + 1/x^2)/lim x^2(2 - 2/x - 1/x^2)

After simplifying the similar terms, we'll get:

(lim 1 + lim 2/x + lim 1/x^2)/(lim 2 - lim 2/x - lim 1/x^2)


Educator Approved

Educator Approved
tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

The limit `lim_(x->oo)(x^2+2x+1)/(2x^2-2x-1)` has to be determined.

Substituting `x = oo` in the given expression `(x^2+2x+1)/(2x^2-2x-1)` gives the result `oo/oo` which is indeterminate. If the result obtained while determining limits is of the form `oo/oo` or `0/0` it is possible to use l'Hospital's rule and substitute the numerator and denominator by their derivatives.

The derivative of (x^2+2x+1) is 2x + 2 and the derivative of (2x^2-2x-1) is 4x - 2.

The given limit can be written as `lim_(x->oo) (2x + 2)/(4x - 2)`

If we substitute `x = oo` , we again get the indeterminate form `oo/oo` . Continue the earlier step and again substitute the denominator and numerator with their derivative.

This gives 2/4 = 1/2

The required limit `lim_(x->oo)(x^2+2x+1)/(2x^2-2x-1) = 1/2`

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the lt(x^2+2x+1)/(2x^2-2x-1) as x-->+infinty.


We divide term by term  both numerator and denominator by x^2 and then take the limit. So.

{ lt(x^2+2x+1)/(2x^2-2x-1) as x-->+infinty} =

lt {1+2x/x^2+1/x^2)/(2-2x/x^2-1/x^2)} as x--> infinity.

Lt(1+2/x-1/x^2)/(2-2/x-1/x^2) as x--> infinity

= (1-2*0+0)/(2-2*0-0), 1/x and 1/x^2 approach zero when x-->inf.



We’ve answered 317,736 questions. We can answer yours, too.

Ask a question