pprove that for alll there is x>0 so
i have tried
limit of f(x) x->infinity = zero
limit of f(x) x->0 = pi/2
so in similar problems i have defined
g(x)= 0.5 when x=0
g(x)= f(x) when 0<x<infinity
g(x)=0 when x=infinity
and then i have to point one is 0 the other is 1/2
and i do mean value theorem
but its not legat to definne such function
infinity is not a point,what to do?
The first limit is correct; the second one should be 1/2. Thus, for each 0 < c < 1/2, you can find x1 > 0 such that c <= f(x1) < 1/2 and x2 > 0 such that f(x2) <= c. The function is continuous for x > 0, so by the intermediate value theorem, there exists an x1 <= x <= x2 such that f(x) = c.
yes i had a typo.
limit of f(x) x->infinity = zero
limit of f(x) x->0 = 1/2
my problem is how to explain that there is such x1 c <= f(x1) < 1/2
from the limit?
same thing goes for
f(x2) <= c
i dont know how to get to these to innequalitties from limit definition
from the first i got
for every x>M -epsilon<f(x)<epsilon
from the second limit definition i get
for every |x-0|<delta 1/2-epsilon<f(x)<epsilon+1/2
i dont know how from these limit definition i get these innequlities
?
1/2-epsilon<f(x)<epsilon+1/2
take epsilon = 1/2 - c
so i get
1+c<f(x)<1-c
-epsilon<f(x)<epsilon
epsilon=c
so i get
-c<f(x)<c
so from the second i get f(x2)<c
how i get c <= f(x1) < 1/2
from
1+c<f(x)<1-c
why its smaller then 1/2
?
I guess you are concerned about f(x) < 1/2 because I said:
According to the graph, f(x) is monotonically decreasing, so f(x) < 1/2. However, it is not important for the current problem. The important thing is that one can find x1 such that f(x1) > c and x2 such that f(x2) < c.Originally Posted by emakarov