limit{x+(x)^1/2}^1/2/x^2-16, x=4

Question

please, help me out to this limit problem

Answer 1

I'm assuming the function is (sqrt(x+sqrt(x))/(x^2-16). Two things to note: the function can't be defined for x=4, because the denominator would be zero; and the function can't be defined for any negative values of x, because we have a term sqrt(x) and we can't take the real square root of a negative number (I'm assuming we're not dealing with imaginary numbers).

We have two scenarios for values of x:

1. 0<=x<4
2. x>4

In (1) the function becomes negative, and in (2) it becomes positive. x=4 is an asymptote. As x approaches 4 from the "left", that is, just less than 4, the value of the function approaches a large negative multiple of sqrt(6). As x approaches 4 from the "right", just greater than 4, the value of the function approaches a large positive multiple of sqrt(6). (Incidentally, at x=0, the function is zero, and as x gets very large and positive the function approaches zero.) I hope this helps.