how to this limit{x+(x)^1/2}^1/2/x^2-16

Question

how to solve this limit{x+(x)^1/2}^1/2/x^2-16

Answer 1

RONG...RONG

if yu want limit as x gotu sumthun av sum funkshun, yu gotta sae

limit av f(x) as x gotu sumthun

Comments

Not sure what muneepenee is getting at!

Answer 2

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).

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

In scenario (1) the function starts at 0 (when x=0) and becomes more negative, approaching some large negative multiple of sqrt(6). In (2) the function starts at some large positive multiple of sqrt(6) and approaches zero as x gets large. x=4 is an asymptote. I hope this helps.