Can the limit be found without using L'hospital rule?

lim x --> 1, [ √(x) - x² ] ÷ [ 1 - √(x) ] = lim x --> 1, [ √(x) - x² ] [ 1 + √(x) ]÷ [ 1 - √(x) ][ 1 + √(x) ]

= lim x --> 1,√(x) [ 1 - x√(x)] [ 1 + √(x) ]÷ (1 - x) = lim x --> 1, √(x) [ 1+ √(x) -x√(x) - x²] /(1-x)

=lim x --> 1,√(x){(1-x)(1+x)+(1-x)√(x)}/(1-x) = lim x --> 1,√(x){(1+x)+√(x)} =√(1){(1+1)+√(1)} = 3

u = √x

[ √(x) - x² ] ÷ [ 1 - √(x) ]

= (u - u⁴)/(1-u)

=[u(1 –u)(1 + u + u²)]/(1 –u)

=u(1 + u + u²) (u ≠ 1)

Lim (u → 1)[ u(1 + u + u²)] =2(1 + 1 + 1) = 3.

[u → 1 as x → 1].

.

[ √(x) - x² ] / [ 1 - √(x) ]

let t = √x

∴ x = t²

[ √(x) - x² ] / [ 1 - √(x) ]

= ( t - (t²)² ) / ( 1 - t )

= ( t - t⁴ ) / ( 1 - t )

= t ( 1 - t³ ) / ( 1 - t ) but difference of cubes (1 - t³ ) = ( 1 - t )( 1 + t + t² )

= t ( 1 - t )( 1 + t + t² ) / ( 1 - t ) cancel out the 1 - t

= t ( 1 + t + t² )

= √x ( 1 + √x + x )

∴

lim x➝1 { √x ( 1 + √x + x ) }

= √1 ( 1 + √1 + 1 )

= 3

Let sqrt(x) = u

(u - u^4) / (1 - u) =>

u * (1 - u^3) / (1 - u) =>

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

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

sqrt(x) * (1 + sqrt(x) + x)

x goes to 1

sqrt(1) * (1 + sqrt(1) + 1) =>

1 * (1 + 1 + 1) =>

3