Hi, is it just me or do text books don't explain the Squeeze/Sandwich theorem properly? I'm always unsure of the domain required. Some help would be very much appreciated =]
The domain isn't really much of a problem, so long as the two bounding functions converge there. The sandwich principle says
If a function $\displaystyle f(z)$ is bound between two convergent functions $\displaystyle g(z),h(z)$ that converge to $\displaystyle l_g,l_h$ respectively then $\displaystyle l_f$ is somewhere between $\displaystyle l_g$ and $\displaystyle l_h$ (providing that $\displaystyle f(z)$ is a convergent function). Hence, if $\displaystyle l_g=l_h$ then $\displaystyle l_f=l_g=l_h$
Please observe that, if a function is bound between two functions that do not converge to the same limit, this does not mean that the first function converges.
Take, for example $\displaystyle f(x)=\sin(x)$ and $\displaystyle g(x)=\frac{3x^2+3x}{(2x+1)(x+4)},h(x)=\frac{(1-x)(2+2x^2)}{x^3}$. Clearly $\displaystyle \lim_{n\rightarrow\infty}g(x)=3/2$, $\displaystyle \lim_{n\rightarrow\infty}h(x)=-2$ and $\displaystyle l_h\leq-1\leq\sin(x)\leq 1 \leq l_g$.
Just because $\displaystyle f(x)$ is bound between two convergent series does not mean that it converges
Paul
hrmm...i think i wasn't clearly pointing out the problem . Let's say we try to prove:
lim x-> 0 x.cos(1/x) = 0
squeeze theorem states that: f(x) <= g(x) <= h(x)
and
limf(x) = lim(hx) = L
then limg(x) = L
i don't know where to begin:
i know you must find a domain for which x.cos(1/x) lies in and then take the limits of the left bound and right bound, which would equal to the limit of the function being squeezed.
But, how to find the domains? Thanks for the reply earlier Paul.
I'll try to help you out quickly.
Let's first work with $\displaystyle \cos \frac{1}{x}$
The cos graph will be oscillating between [-1 ; 1]
Therefore:
$\displaystyle -1 \leq \cos \frac{1}{x} \leq 1$
Multiply by x through the inequality.
$\displaystyle -x \leq x \cos \frac{1}{x} \leq x$
If x approaches zero, we obtain the following:
$\displaystyle 0 \leq x \cos \frac{1}{x} \leq 0$
And if the left and right is zero, then the middle must also be zero... (That's logical! )