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Simple Limit Query

I have a function, f(x)=3x−723x2−6x+7f(x) = \dfrac{3x-\frac{7}{2}}{3x^2-6x+7}.

I'm trying to find its asymptotes, and so I am trying to find the limit of f(x) as x approaches +- infinity. Now, I know that it approaches 0 from above the x axis when it approaches +infinity, as the function tends to: 3x3x2−6x→1x−2→1x\frac{3x}{3x^2-6x} \rightarrow \frac{1}{x-2} \rightarrow \frac{1}{x}.

Now I also know that it approaches 0 from below the x axis as it approaches minus infinity. How do I show this, is it as simple as saying as the limit is essentially 1/x, I'm looking at 1/-inf therefore the answer is '-0'?
Reply 1
Avatar for Hasufel
Hasufel
16
the whole function breaks down as: [6x-7/(6x^2-12x+14)]
to examine the behaviour at x= +/- infinity, you divide the function by the highest power of x in the denominator - looking at the fuction in "bits" of a/x or a/x^2 gives the answer.

that`s the informal way of doing it

function doesn`t have any asymptotes - has 2 tangents though
(edited 12 years ago)
Original post by ViralRiver
I have a function, f(x)=3x−723x2−6x+7f(x) = \dfrac{3x-\frac{7}{2}}{3x^2-6x+7}.

I'm trying to find its asymptotes, and so I am trying to find the limit of f(x) as x approaches +- infinity. Now, I know that it approaches 0 from above the x axis when it approaches +infinity, as the function tends to: 3x3x2−6x→1x−2→1x\frac{3x}{3x^2-6x} \rightarrow \frac{1}{x-2} \rightarrow \frac{1}{x}.

Now I also know that it approaches 0 from below the x axis as it approaches minus infinity. How do I show this, is it as simple as saying as the limit is essentially 1/x, I'm looking at 1/-inf therefore the answer is '-0'?

If you're just looking for the limit (without rigorous proof), dividing the numerator and denominator by the highest power of x present in either the numerator or the denominator (which is x^2 in this case) we get f(x)=3x−72x23−6x+7x2f(x)=\dfrac{\frac{3}{x} - \frac{7}{2x^2}}{3-\frac{6}{x}+\frac{7}{x^2}}.

Taking the limit from here should be straightforward.

EDIT: Hasufel got there first.
Reply 3
Avatar for ViralRiver
ViralRiver
OP
16
So the limit = 0? But Surely that means there is an asymptote? But how do I know whether it is limiting 0 from above/below, for +- infinity?
Reply 4
Avatar for ztibor
ztibor
10
Original post by ViralRiver
So the limit = 0? But Surely that means there is an asymptote? But how do I know whether it is limiting 0 from above/below, for +- infinity?


Independently from 'above' or 'below' the limit is 0 so the equation of the
horizontal asymptote is
f(x)=0f(x)=0 or y=0y=0 ('both from above an bellow').
For example for the 1x2\frac{1}{x^2} the limit is 0 from above
both for + infinity or - infinity and the horizontal asymptote is y=0.
However calculating in Re=R∪(−∞;+∞)\mathbb R_e=\mathbb R \cup \left (-\infty; +\infty \right ) for c>0
c+∞=0+\frac{c}{+\infty}=0^{+} (above)
c−∞=0−\frac{c}{-\infty}=0^{-} (below)
for c<0
c+∞=0−\frac{c}{+\infty}=0^{-} (below)
c−∞=0+\frac{c}{-\infty}=0^{+} (above)
otherwise f. e for 1x−2\frac{1}{x-2}
lim⁡x→2−=12−2−=10−=−∞\lim_{x \rightarrow 2^{-}}=\frac{1}{2-2^{-}}=\frac{1}{0^{-}}=-\infty (from the left)
lim⁡x→2+=12−2+=10+=+∞\lim_{x \rightarrow 2^{+}}=\frac{1}{2-2^{+}}=\frac{1}{0^{+}}=+\infty (from the right)
So the x=2x=2 line is a vertical asymptote here.
Your original f(x) has no vertical asymptote.
General rule for asymptote with equation of ax+bax+b
a=lim⁡x→∞f(x)xa=\lim_{x \rightarrow \infty} \frac{f(x)}{x}
b=lim⁡x→∞(f(x)−a⋅x)b=\lim_{x \rightarrow \infty} (f(x)-a\cdot x)
(edited 12 years ago)

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