t has no limit. It will go to infinity.

lim x-->∞ (x-sqrt(x^2+x))

multiply and divide bby (x+sqrt(x^2+x))

= lim x-->∞ (x-sqrt(x^2+x))(x+sqrt(x^2+x)) /(x+sqrt(x^2+x))

The numerator is of the form (a-b)(a+b)=a^2-b^2

a= x

a^2 = x^2

b= sqrt(x^2+x)

b^2=x^2+x

a^2-b^2 = x^2-(x^2+x) = -x

= lim x-->∞ -x /(x+sqrt(x^2+x))

apply L'Hopital's rule

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

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

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

lim x-->∞ (2x+1) /(2sqrt(x^2+x))

= lim x-->∞ x(2+1/x) / (2x(sqrt(1+1/x))

= lim x-->∞ (2+1/x) /(2sqrt(1+1/x)) = 1 (since 1/x approaches 0)

= -1 /(1+1)

= -1/2

x - sqrt(x^2 + x) =>

(x - sqrt(x^2 + x)) * (x + sqrt(x^2 + x)) / (x + sqrt(x^2 + x)) =>

(x^2 - (x^2 + x)) / (x + sqrt(x^2 + x)) =>

(x^2 - x^2 - x) / (x + sqrt(x^2 + x)) =>

-x / (x + sqrt(x^2 + x)) =>

-x / (x + x * sqrt(1 + 1/x)) =>

-1 / (1 + sqrt(1 + 1/x))

Do you still need L'hopital?

x goes to infinity

-1 / (1 + sqrt(1 + 1/inf)) =>

-1 / (1 + sqrt(1 + 0)) =>

-1 / (1 + sqrt(1)) =>

-1 / (1 + 1) =>

-1/2

x - sqrt(x^2 + x) =>

x - x * sqrt(1 + 1/x) =>

x * (1 - sqrt(1 + 1/x)) =>

(1 - sqrt(1 + 1/x)) / (1/x)

x goes to infinity

u = 1/x

u goes to 0

(1 - sqrt(1 + u)) / u

f(u) = 1 - sqrt(1 + u)

f'(u) = (-1/2) * (1 + u)^(-1/2)

g(u) = u

g'(u) = 1

(-1/2) * (1 + u)^(-1/2) / 1 =>

-1 / (2 * (1 + u)^(1/2))

u goes to 0

-1 / (2 * (1 + 0)^(1/2)) =>

-1 / (2 * 1) =>

-1/2

The limit does not exist.