How do you find the asymptotes for #g(x)=x/(root4(x^4+2))#?

Question

Lily Anderson · Accepted Answer

Evaluate the limits of #g(x)# as #x->+oo# and #x->-oo# to find horizontal asymptotes #y=1# and #y=-1#.

#g(x) = x/root(4)(x^4+2) = x/abs(x) abs(x)/root(4)(x^4+2) = x/abs(x) 1/root(4)(1+2/x^4)# So #lim_(x->+oo) g(x) = 1# and #lim_(x->-oo) g(x) = -1# So #g(x)# has horizontal asymptotes #y=1# and #y=-1# #x^4+2 >= 2 > 0# for any Real number #x# Hence the denominator of #g(x)# is always non-zero and #g(x)# has no vertical asymptotes. graph{x/root(4)(x^4+2) [-5.55, 5.55, -2.775, 2.774]}

Greyson Brown · Answer

undefined To find the asymptotes of the function $g(x) = \frac{x}{\sqrt[4]{x^4 + 2}}$, you need to consider both horizontal and vertical asymptotes.

1. Vertical Asymptotes:
   Vertical asymptotes occur where the denominator becomes zero. However, in this case, the denominator is a fourth root, and the expression inside the root, $x^4 + 2$, is always positive for real $x$ because any real number raised to an even power is non-negative, and adding 2 to it ensures it's always positive. Therefore, there are no vertical asymptotes for this function.

2. Horizontal Asymptotes:
   To find horizontal asymptotes, analyze the behavior of the function as $x$ approaches positive or negative infinity.
   As $x$ approaches infinity, $x$ dominates the expression. Since both the numerator and the denominator approach infinity, the function has a horizontal asymptote at $y = 1$, as the term inside the root becomes insignificant compared to $x$.

Therefore, the horizontal asymptote of the function $g(x)$ is $y = 1$, and there are no vertical asymptotes.