What are the asymptotes and removable discontinuities, if any, of #f(x)= sqrt(x)/(e^x-1)#?

1 Answer
George C.
Jun 16, 2017

#f(x)# has a horizontal asymptote #y=0# and a vertical asymptote #x=0#

Explanation:

Given:

#f(x) = sqrt(x)/(e^x-1)#

  • The domain of the numerator #sqrt(x)# is #[0, oo)#

  • The domain of the denominator #e^x - 1# is #(-oo, oo)#

  • The denominator is zero when #e^x = 1#, which for real values of #x# only occurs when #x=0#

Hence the domain of #f(x)# is #(0, oo)#

Using the series expansion of #e^x#, we have:

#f(x) = sqrt(x)/(e^x - 1)#

#color(white)(f(x)) = sqrt(x)/((1+x+x^2/2+x^3/6+...) - 1)#

#color(white)(f(x)) = sqrt(x)/(x+x^2/2+x^3/6+...)#

#color(white)(f(x)) = 1/(sqrt(x)(1+x/2+x^2/6+...)#

So:

#lim_(x->0^+) f(x) = lim_(x->0^+) 1/(sqrt(x)(1+x/2+x^2/6+...))#

#color(white)(lim_(x->0^+) f(x)) = lim_(x->0^+) 1/(sqrt(x)(1+0+0+...))#

#color(white)(lim_(x->0^+) f(x)) = lim_(x->0^+) 1/(sqrt(x))#

#color(white)(lim_(x->0^+) f(x)) = +oo#

and:

#lim_(x->oo) f(x) = lim_(x->oo) 1/(sqrt(x)(1+x/2+x^2/6+...)) = 0#

So #f(x)# has a vertical asymptote #x=0# and a horizontal asymptote #y=0#

graph{sqrt(x)/(e^x-1) [-6.1, 13.9, -2.92, 7.08]}

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