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June 27, 2016
Posted by **kim** on Thursday, December 10, 2009 at 8:00pm.

f(x)=x/(x-1)

q(x)=(x-1)/x

- pre calc -
**bobpursley**, Thursday, December 10, 2009 at 8:06pmWhen the denominator equals zero, there is a vertical asymptoe

When the numerator becomes constant as x approaches very large, there is a horizontal asymptoe.

For instance, in the second

q(x)=(x-1)/x= 1-1/x as x gets very large, it becomes 1-0 or 1

In the first,

f(x)=x/(x-1)= 1/(1-1/x) (multiplied numerator and denominator by 1/x)

f(x)= 1/(1-0) when x is very large, or f(x)=1 - pre calc -
**anonymous**, Thursday, December 10, 2009 at 8:08pmx = 1 for the first one

x = 0 for the second - pre calc -
**bobpursley**, Thursday, December 10, 2009 at 8:09pmyes, for the vertical asy.

- pre calc -
**kim**, Thursday, December 10, 2009 at 8:15pmso what about this one g(x)=1.5^x

- pre calc -
**bobpursley**, Thursday, December 10, 2009 at 8:19pmNo vertical, no horizontal.

- pre calc -
**kim**, Thursday, December 10, 2009 at 8:21pmhow can you tell it has neither

- pre calc -
**MathMate**, Thursday, December 10, 2009 at 11:34pmAs Mr. Bob mentioned:

When the denominator becomes zero when x=c and c is finite, there is a vertical asymptote.

A horizontal asymptote is typically identified by the fact that lim f(x) approaches a constant value as x->∞ or -> -&infin.

In the case of:

g(x)=1.5^x

there is no denominator that makes g(x) infinite when x is finite, so no vertical asymptote.

g(x) becomes infinite when x->+∞, so no horizontal asymptote on the right. But on the left..., as x->-∞, g(x) approaches zero, so what do you think?