Evaluate the limit . lt[(x+5)^2-x^2]/(2x+14). x-->infinite

Expert Answers
hala718 eNotes educator| Certified Educator

limit f(x) n= limit [(x+5)^2 - x^2]/(2x+14)  x--> inf

First let us open brackets:

==> lim f(x) = limit (x^2 + 10x + 25 - x^2) /(2x+14)  

                    = limit ( 10x+25)/(2x+14)

                    = limit 5(2x+5)/2(x+7)

                   = (5/2) limit (2x+5)/(x+7)

Now let is divide by heighet power (x):

==> limit f(x) = (5/2)* limit (2+5/x)/(1+7/x)

When x --> inf:

==> limit f(x) = (5/2) * 2/1 = 5

Then the limit = 5

neela | Student

To evaluate the limit . lt[(x+5)^2-x^2]/(2x+14). x-->infinite

We factor the numerator .

Numerator = (x+5)^2-x^2 = (x+5+x)(x+5-x) as a^2-b^2 = (a+b)(a-b).

Numerator = (2x+5)(5)

Denominator = 2x+14

Thefreore lt ({(x+5)^2-x^2}/(2x+14) = Lt {5(2x+5)/(2x+14)}

= lt (5(2+5/x)/(2+14/x as x-->ifinity.

= 5(2 +0)/(2+0) , as 5/x =0 and 14/x = 0 ax x--> infinity.

= 10/2 = 5.

Lt{ (x+5^2)^2 -x^2}/(2x+14) = 5, as x--> infinity.

giorgiana1976 | Student

If x tends to +inf.,we'll evaluate the limit of the function by factorizing both, numerator and denominator, by the highest power of x.

For the beginning, we'll expand the square from numerator:

(x+5)^2 = x^2 + 10x + 25

Now, we'll remove the brackets from numerator:

(x+5)^2-x^2 = x^2 + 10x + 25 - x^2

We'll eliminate like terms and we'll get:

(x+5)^2-x^2 = 10x + 25

We'll re-write the limit:

lt[(x+5)^2-x^2]/(2x+14) = lt (10x + 25) / (2x+14)

We'll factorize the numerator by 5x and denominator by 2x:

lt (10x + 25) / (2x+14) = lt 5x(2 + 5/x) / 2x(1 + 7/x)

We'll reduce the similar terms:

 lt 5x(2 + 5/x) / 2x(1 + 7/x) =  lt 5(2 + 5/x) / 2(1 + 7/x)

After reducing similar terms,we'll get:

 lt 5(2 + 5/x) / 2(1 + 7/x) = 5* (2+0)/2*(1+0) = 10/2

 lt 5(2 + 5/x) / 2(1 + 7/x) = 5