Question

Question #7c54f

Answer 1

see below

Explanation:

`Lim_(xrarr0)sqrt(x²+2x+2)-x=sqrt2`

or

`Lim_(xrarroo)sqrt(x²+2x+2)-x*(sqrt(x²+2x+2)+x)/(sqrt(x²+2x+2)+x)`

`Lim_(xrarroo)(x²+2x+2-x^2)/(sqrt(x²+2x+2)+x)`

`Lim_(xrarroo)(2x+2)/(sqrt(x²+2x+2)+x)`

`Lim_(xrarroo)(cancelx(2+2/x))/(cancelx(sqrt(1+2/x+2/x^2)+1)`

`Lim_(xrarroo)(2+2/x)/(sqrt(1+2/x+2/x^2)+1)=2/(1+1)=1`

You have a mistake but a correct solution :D

Answer 2

You have the basic ideas but there are some errors in your details (and these errors canceld out so you got the correct answer.)

Explanation:

`((sqrt(x²+2x+2)-x))/1 * ((sqrt(x²+2x+2)+x))/((sqrt(x²+2x+2)+x)) = ((x^2+2x+2)-x^2)/(sqrt(x²+2x+2)+x)`

`= (2x+2)/(sqrt(x^2(1+2/x+2/x^2))+x)` `" "` for `x != 0`

`= (2x+2)/(sqrt(x^2)sqrt(1+2/x+2/x^2)+x)`

Use `sqrt(x^2) = absx` and `x > 0` (so `absx=x`) to get

`= (2x+2)/(xsqrt(1+2/x+2/x^2)+x)`

`= (x(2+2/x))/(x(sqrt(1+2/x+2/x^2)+1))`

`= (2+2/x)/(sqrt(1+2/x+2/x^2)+1)`

Now find the limit (I assume we are finding the limit at infinity)

`lim_(xrarroo)(sqrt(x²+2x+2)-x) = lim_(xrarroo)(2+2/x)/(sqrt(1+2/x+2/x^2)+1)`

`= (2+0)/(sqrt(1+0+0)+1) = 2/2=1`

Note that
Your `2x^2` in the numerator is incorrect

and

`lim_(xrarr00)(2x+2+(2/x))/(sqrt(1+(2/x)+(2/x²))+1) = (oo+2+0)/(sqrt1+1) = oo` `" "`

The limit of your expression is not `" "` `2/2=1`