- #1
CallMeShady
- 45
- 1
Finding Limit As "X" Approaches Infinite Of Square Root Function
None that I am aware of.
What I tried to do to solve this problem was first, multiplying the function by its conjugate, and then simplifying the numerator which yielded "5x" divided by the original conjugate. Then I factored out "x-squared" from the denominator and after taking the square root of that, I got the absolute value of "x". Since the limit approaches positive infinity, the absolute value of "x" becomes just "x" by itself. This divided out the "x's" from the numerator and the denominators. Then I simply substituted positive infinity for "x's" that remained in the square roots in the denominators and got the answer of 5/2... which is apparently wrong. Can anyone help please?
Homework Statement
Homework Equations
None that I am aware of.
The Attempt at a Solution
What I tried to do to solve this problem was first, multiplying the function by its conjugate, and then simplifying the numerator which yielded "5x" divided by the original conjugate. Then I factored out "x-squared" from the denominator and after taking the square root of that, I got the absolute value of "x". Since the limit approaches positive infinity, the absolute value of "x" becomes just "x" by itself. This divided out the "x's" from the numerator and the denominators. Then I simply substituted positive infinity for "x's" that remained in the square roots in the denominators and got the answer of 5/2... which is apparently wrong. Can anyone help please?