Answer to this limit without using L'Hospital's Rule?

Question

lim x->1 [(3*(x^2) - 4)*sq(x) + 1]/[x - 1]

I know the answer is 5.5 but how can you get this limit without using L'Hospital's Rule?

Answer 1

it's

l'Hôpital's rule

not hospital

try factoring instead.....

((3x^2 -4)* x^.5 +1) / (x-1) =
((3x^2 -4 +x -x)* x^.5 +1) / (x-1) = {notice I added/subt. x?}
((3x^2 +x -4 -x)* x^.5 +1) / (x-1) = {rearranged x}
(([3x^2 +x -4] -x)* x^.5 +1) / (x-1)=
([(x-1)*(3x+4)] -x)* x^.5 +1) / (x-1) = {I factored the 3x^2... term}

([(x-1)*(3x+4)] -x)* x^.5 +1) / (x-1)

([(x-1)*(3x+4)]*x^.5 - x* x^.5 +1) / (x-1)

[(x-1)*(3x+4)]*x^.5 / (x-1) - [(x^3/2 - 1) / (x-1)]

= (3x+4)*x^.5 - [(x^3/2-1) / (x-1)]

but since (a^3 - b^3) = (a-b) x (a^2 + ab+b^2) and since 1^3 = 1, and x^1/2 cubed = x^3/2.....

= (3x+4)*x^.5 - [(x^1/2-1)*(x+x^1/2+1)] / (x-1)

= (3x+4)*x^.5 -(x+x^1/2+1) * ( x^1/2-1)/(x-1)

but

(x-1) = (x^1/2 - 1)* (x^1/2 +1)

and substituting.....

= (3x+4)*x^.5 -(x+x^1/2+1) * ( x^1/2-1)/[(x^1/2 - 1)* (x^1/2 +1)]

and canceling.....

= (3x+4)*x^.5 -(x+x^1/2+1) / (x^1/2 +1)

now plug in the value of x = 1....

= 7 - (3) / (2) = 5.5

Answer 2

multiply and divide by (3x^2-4) sqrt(x)-1 numerator and denominator
you get
[(3x^2-4)^2*x-1]/(x-1) *1/[3x^2-4)sqrt(x)-1]
The first part can be factored as
(x-1)(9x^4+9x^3-15x^2-15x+1) and the 2nd factor==>-1/2
so simplifying (x-1) the limit is 11/2=5.5
The first answer is nothing but L´Hôpital.

Answer 3

lim when x approaches a of (f(x) - f(a)) / (x - a) is f '(a)

then f(x) = (3x² - 4)sqrt(x) and a = 1 f(a) = -1

f '(x) = 6x sqrt(x) + (3x² - 4) / (2sqrt(x))

f '(1) = 6 + (-1)/2 = 11/2 = 5.5

Answer 4

5.5