in the graph y=lnx/x, why is y = 0 an assmptote?

Question

my teacher's answers say that y=0 is an assymptote, but when i sub y=0 in to the equation, i get x=1. is this possible? and why?

Answer 1

Some people think of an asymptote as a curve (often a horizontal or vertical line) that a function approaches as the x-values approach a certain value (number or infinity).

For your function, the domain of y=ln(x)/x is (0,\infty). As x approaches zero from the right ln(x) approaches -\infty and 1/x approaches +\infty so ln(x)/x approaches negative infinity as x approaches 0 from the right. In other words, the vertical line x=0 (the y-axis) is a vertical asymptote.

As x approaches +\infty, both ln(x) and x approach \infty so ln(x)/x is called an "indeterminate form of type \infty/\infty." Use can use L'Hospital's rule to find the limit.

L'Hospital's rule says that if f(x)/g(x) is an indeterminate form of type 0/0 or \infty/\infty if x=a (where a may be \infty), then \lim_{x \to a} f(x)/g(x)=\lim_{x \to a} f'(x)/g'(x).

Thus, in this case,

\lim_{x \to \infty} ln(x)/x=
\lim_{x \to \infty} (1/x)/1=
\lim_{x \to \infty} 1/x=0.

This means that as x approaches infinity, the corresponding y-values approach 0. That is, the horizontal line y=0 (the x-axis) is a hoizontal asymptote.

Of course, you are correct that if y=0, then x=1. This shows you that (1,0) is an x-intercept for a function.

It also shows you that a function can cross its asymptotes.

You can see this further by looking at a function like y=sin(x)/x. y=0 if x=n \pi, where n represents any integer. On the other hand, the line y=0 is also a horizontal asymptote.

Answer 2

Y is a function of X, not the other way around. You can't substitute for Y and get a meaningful answer. You have to substitute for X instead.

Since X is a denominator in this case, you can't substitute 0 for X without getting an undefined result; so, you have to use a LIMIT instead of a subsitution.

Note:

The LIMIT of lnx/x as x goes to 0 is

ln(0)/0 = 1/0 = infinity.

Thus, as X goes to 0,

Y goes to infinity.

Y = 0 is the asymptote, because X can never get to 0 without making Y go to infinity.

Another way to look at it: the curve, as X goes to 0, begins to climb steeply along the Y axis to Y = infinity. The curve cannot ever touch the Y axis because X can never get to 0, so the Y axis (or Y = 0) is the asymptote.

Answer 3

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Answer 4

The y axis is a 'vertical' asymptote (and note the spelling of 'asymptote ) since lnx -> -∞ as x -> 0. At x = 1 the function has the value of 0/1 or 0.

However..... As x -> ∞ the value of lnx/x -> 0 so that y=0 (the horizontal x axis) is an asymptote as x -> ∞.

Hope that helps.


Doug

YO!! almintaka. ln(0) is **undefined**. And the range of ln(x) over the domain 0

Answer 5

My original answer was wrong... its too late.

You are correct that Y=0 gives you x=1 but if you graph it out... you'll see that it never gets back to Y=0 again... y=0 is an asymptote in the range 1
I don't agree with the other posts. You sure can substitute 0 in for y regardless of what's a function of what. That's the beauty of algebra... if it didn't work both ways... we'd be screwed.

Answer 6

Your teacher is wrong or you understood him/her wrong.
if y= ln x / x
and y=0, then ln x = 0
taking exponentials of both sides x = e^0 = 1
however x=0 is an asymptote, consider x=0, then y= ln 0 / 0 , division by 0 is undefined but for this purpose you can say that it's an asymptote, end of confusion.
Talk to your teacher about this, hope this helped.

Answer 7

The function is defined only for x>0

The limit of lnx/x when x goes to zero by right side(the only possible) is infinity

So x=0 is a vertical asymptote

Answer 8

i dont know y