determine if it exists Lim x^2-2x+1/sqrt(x+3)-2
x->1

Lim x^2-2x+1 *sqrt(x+3 ) +2 / (sqrt(x+3)-2)(sqrt(x+3)+2)
x->1

Lim (x-1)(x-1)*sqrt(x+3 ) +2 /(x-1)
x->1

Lim (x-1) sqrt(x+3 ) +2 = 0
x->1

2. Originally Posted by bobby77
determine if it exists Lim x^2-2x+1/sqrt(x+3)-2
x->1

Lim x^2-2x+1 *sqrt(x+3 ) +2 / (sqrt(x+3)-2)(sqrt(x+3)+2)
x->1

Lim (x-1)(x-1)*sqrt(x+3 ) +2 /(x-1)
x->1

Lim (x-1) sqrt(x+3 ) +2 = 0
x->1
Hello,

Greetings

EB

determine if it exists Lim (x^2-2x+1)/(sqrt(x+3)-2)
x->1

Lim (x^2-2x+1) *(sqrt(x+3 ) +2) /[ (sqrt(x+3)-2)(sqrt(x+3)+2) ]
x->1

Lim [(x-1)(x-1)]*(sqrt(x+3 ) +2) /(x-1)
x->1

Lim (x-1)*( sqrt(x+3 ) +2) = 0
x->1

4. Originally Posted by bobby77
determine if it exists Lim (x^2-2x+1)/(sqrt(x+3)-2)
x->1

Lim (x^2-2x+1) *(sqrt(x+3 ) +2) /[ (sqrt(x+3)-2)(sqrt(x+3)+2) ]
x->1

Lim [(x-1)(x-1)]*(sqrt(x+3 ) +2) /(x-1)
x->1

Lim (x-1)*( sqrt(x+3 ) +2) = 0
x->1
Or more simply ya got,
$\lim_{x\to 1}\frac{(x-1)^2}{\sqrt{x+3}-2}$
Rationalize,
$\frac{(x-1)^2}{\sqrt{x+3}-2}\cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}$
Thus,
$\frac{(x-1)^2(\sqrt{x+3}+2)}{x-1}$
Thus,
$(x-1)(\sqrt{x+3}+2)$
As, $x\to 1$ you get 0.