#Limits help needed
54 messages · Page 1 of 1 (latest)
it should be 1
what are you allowed to use
how
trig identitities
I think you can use the squeeze theorem
ohhhh
how would we use squeeze theorem on that anyway
we would use other functions 💀
wait
bludicus
my fault
yes you use the squeeze theorem
but you can only use trig identities
right?
that would be prefferable
ok
no these functions don't apply to the squeeze theorem
because it states that if f(x)<=g(x)<=h(x)
and f(p)=h(p)
also you can use 1 as the upper bound but you have to use that |sin x| ≤ |x| for all x
but in here
the function that is supposed to be less than sin(x)/x is greater at one point
so you will need a different function
use taylor series expansion of sin
or squeeze sin(x) between x and x-x^3/6
what are you talking about
the limit of both functions as they approach 0 is 1
and their function values at 0 are 1
at other micro values around 0 and in general, it's either too big or too small
the function that is supposed to be less than g(x)
which in this case is sin(x)/x
it is greater than g(x) at some point
but it has to be
less than or equal to
or I may be wrong 🤷♂️
i remember it being equal to g(x) at one point but >= or <= at all the other points of the respective graphs
ngl I don't think it's possible w/o squeeze theorem
to be sure, are you only allowed to use squeeze theorem & trig identities? see if you can use l'hopital or taylor series
Actually, it's solved historically way before l'hopitale's and it's various proofs include a lot of the tools used to prove l'hopitale
there is a geometric proof too
n-gons-->circles
update
it turns out you can use squeeze theorem
or just actually find the limit
by approximation
how do you end the question
.solved