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Asymptote

y=1/x

how do we prove their is a horizontal asymptote

this one is kind of easy as obviously 1≠0 but say you had something a lot more complicated - how would you go about that?
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Gaz031
15
You can read the definition of an asymptote at the following link: http://www.math.com/tables/derivatives/extrema.htm

We want to prove that y=0 is a horizontal asymptote and so need to show that 1x\frac{1}{x} tends to 0 as x tends to plus or minus infinity. You can clearly see this is the case but probably can't prove it rigorously at A-Level. I'll show you how you could do it for x tending to infinity, the proof for x tending to minus infinity is very similar.

We say that a function f(x)f(x) tends to a limit ll as xx tends to infinity if (and only if) for every ϵ>0\epsilon > 0 there exists XX such that ∣f(x)−l∣<ϵ|f(x)-l|<\epsilon for all x>Xx>X
This definition will probably look intimidating but after thought and a few examples you'll probably see it makes sense - you can make ϵ\epsilon as small as you like so the difference between f(x) and l becomes arbitrarily small. Generally smaller values of ϵ\epsilon give rise to larger values of X.

For the example in which we wish to prove that 1x→0\frac{1}{x} \rightarrow 0 as x→∞x \rightarrow \infty,
∣1x−0∣=∣1x∣=1x<ϵ|\frac{1}{x}-0|=|\frac{1}{x}|=\frac{1}{x}<\epsilon for all x>1ϵx>\frac{1}{\epsilon} (for the case where x>0).
Hence set X=1ϵX=\frac{1}{\epsilon} Then ∣1x−0∣<ϵ|\frac{1}{x}-0|<\epsilon for all x>Xx>X and so 1x→l\frac{1}{x} \rightarrow l as x→∞x \rightarrow \infty

Don't worry if that flew over your head... you don't need to worry about it unless you do something mathematical at university - it just shows that the process of functions 'approaching' certain values can be rigorously investigated.

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