lim (√ x² + 2x) - x
x→∞
* (x² + 2x) is all under the sq root
* answer is "1"
x→∞
* (x² + 2x) is all under the sq root
* answer is "1"
-
You have to multiply by its conjugate : which is squareroot(x^2+2x) + x
= x^2 + 2x - x^2 / squareroot(x^2+2x) +x
Now you have to divide by the highest power from the denominator.
= x^2 + 2x - x^2 / |x| squareroot( x^2/x^2+2/x))+ x/x <== be care ful here the square root of x^2 is |x|
= x( x^2/x + 2x/x - x^2/x)/ ^
= Since absolute value of + infinity is positive you leave it as positive if it were -infinity you change to a -1 when crossing out
= x + 2 - x / squareroot(1 + 2/x ) +1
= 0 + 2 - 0/ squareroot(1+0) + 1
= 2 / 1 +1= 2/2 = 1
= x^2 + 2x - x^2 / squareroot(x^2+2x) +x
Now you have to divide by the highest power from the denominator.
= x^2 + 2x - x^2 / |x| squareroot( x^2/x^2+2/x))+ x/x <== be care ful here the square root of x^2 is |x|
= x( x^2/x + 2x/x - x^2/x)/ ^
= Since absolute value of + infinity is positive you leave it as positive if it were -infinity you change to a -1 when crossing out
= x + 2 - x / squareroot(1 + 2/x ) +1
= 0 + 2 - 0/ squareroot(1+0) + 1
= 2 / 1 +1= 2/2 = 1