means to say that for any given , we can find
such that anytime
(i.e. the whenever
is "close enough" to 5), we can guarantee that
(i.e. the value of
is "close enough" to the limit value).
What we want to end up with is
Dividing both sides by 3 gives
which suggests is a sufficient threshold.
The proof itself is essentially the reverse of this analysis: Let be given. Then if
and so the limit is 7. QED
Step-by-step explanation:
The frequency of sine function is given by the number of periods in a given range. For example:
This means that, if we want another sine function with frequency 20% higher, we need that function to have a frequency of 1.2 in the interval .
To be easier to see we will consider interval instead of
. In this interval
has 5 periods, therefore our new sine function should have 6 periods.
Finally, as we can see in the graph, the function (in blue) has a frequency 20% higher than
(in red). This can be easily seen counting the number of periods between 0 and
for both functions. 5 for
and 6 for
.
