Definition 6
Let f(x)f(x) be a function defined on x>Kx>K for some KK. Then we say
that,limx→∞f(x)=Llimx→∞⁡f(x)=L
if for every number ε>0ε>0 there is some number M>0M>0 such that
|f(x)−L|<εwheneverx>M|f(x)−L|<εwheneverx>M
Definition 7
Let f(x)f(x) be a function defined on x<Kx<K for some KK. Then we say
that,limx→−∞f(x)=Llimx→−∞⁡f(x)=L
if for every number ε>0ε>0 there is some number N<0N<0 such that
|f(x)−L|<εwheneverx<N|f(x)−L|<εwheneverx<N
To see what these definitions are telling us here is a quick sketch
illustrating Definition 6. Definition 6 tells us is that no matter how close
to LL we want to get, mathematically this is given by |f(x)−L|<ε|f(x)−L|<ε for
any chosen εε, we can find another number MM such that provided we
take any xx bigger than MM, then the graph of the function for that xx will
be closer to LL than either L−εL−ε and L+εL+ε. Or, in other words, the
graph will be in the shaded region as shown in the sketch below.
Finally, note that the smaller we make εε the larger we’ll probably need to
make MM.
Here’s a quick example of one of these limits.
Example 6 Use the definition of the limit to prove the following
limit.limx→−∞1x=0limx→−∞⁡1x=0
Hide Solution
Let ε>0ε>0 be any number and we’ll need to choose a N<0N<0 so that,
∣∣∣1x−0∣∣∣=1|x|<εwheneverx<N|1x−0|=1|x|<εwheneverx<N
Getting our guess for NN isn’t too bad here.
1|x|<ε⇒|x|>1ε1|x|<ε⇒|x|>1ε
Since we’re heading out towards negative infinity it looks like we can
choose N=−1εN=−1ε. Note that we need the “-” to make sure that NN is
negative (recall that ε>0ε>0).
Let’s verify that our guess will work. Let ε>0ε>0 and
choose N=−1εN=−1ε and assume that x<−1εx<−1ε. As with the previous
example the function that we’re working with here suggests that it will be
easier to start with this assumption and show that we can get the left
inequality out of that.
x<−1ε|x|>∣∣∣−1ε∣∣∣take the absolute value|x|>1εdo a little
simplification1|x|<εsolve for |x|∣∣∣1x−0∣∣∣<εrewrite things a
littlex<−1ε|x|>|−1ε|take the absolute value|x|>1εdo a little
simplification1|x|<εsolve for |x||1x−0|<εrewrite things a little
Note that when we took the absolute value of both sides we changed both
sides from negative numbers to positive numbers and so also had to
change the direction of the inequality.
So, we’ve shown that,
∣∣∣1x−0∣∣∣=1|x|<εwheneverx<−1ε|1x−0|=1|x|<εwheneverx<−1ε
and so by the definition of the limit we have,
limx→−∞1x=0