How do you use limit approach to identify all asymptotes of the function f(x)=(1/x)+ln(x)?

Answer 1

A function f(x) has a horizontal asymptote at +∞, provided

lim(f(x)) as x->∞ exists and is neither +∞ nor -∞.

A function f(x) has a horizontal asymptote at -∞, provided

lim(f(x)) as x->-∞ exists and is neither +∞ nor -∞.

A function f(x) has a vertical asymptote at x=a, provided

+∞ = lim(f(x)) as x->a-, or

-∞ = lim(f(x)) as x->a-, or

+∞ = lim(f(x)) as x->a+, or

-∞ = lim(f(x)) as x->a+

Your function f(x) is not defined for x < 0, so there is no horizontal asymptote at -∞.

The only potential value of a, where f(x) could have a vertical asymptote is a=0.

So we just need to evaluate

lim(f(x)) as x->+∞ and as x->0+.

For both cases, it is easier to first compute

lim(e^f(x)) = lim(e^1/x x)

because exponentiation is monotone increasing.

1) As x->+∞,

lim(e^1/x) = 1, and

lim(x) = +∞

Therefore lim(e^1/x x) = +∞

Therefore lim(f(x)) = +∞

2) As x->0+

lim(e^1/x x) = lim(e^1/x / (1/x))

As x->0+, both numerator as denominator go towards ∞.

Therefore we can apply the L'Hospital Rule.

lim(e^1/x / (1/x)) =

lim(e^1/x(-1/x²) / (-1/x²)) =

lim((e^1/x) = ∞

Therefore lim(f(x)) = ∞

The conclusion is that f(x) has no horizontal asymptote,

but has vertical asymptote at x=0.