i made a limit on both infinity and minus infinity for them
and i tried to find but its not working
http://img201.imageshack.us/img201/5458/23597303em5.gif
i made a limit on both infinity and minus infinity for them
and i tried to find but its not working
http://img201.imageshack.us/img201/5458/23597303em5.gif
The domain of the function $\displaystyle \mathrm{ln}(x)$ is $\displaystyle (0,\infty)$, with $\displaystyle \lim_{x\to0}\mathrm{ln}(x)=-\infty$.
Let
$\displaystyle f(x):=\frac{1}{1+(\mathrm{ln}(x))^{2}},$
then
$\displaystyle f^{\prime}(x):=-\frac{2\mathrm{ln}(x)}{x(1+(\mathrm{ln}(x))^{2})^{ 2}},$
which implies $\displaystyle f^{\prime}(1)=0$.
Therefore its maximum value is $\displaystyle f(1)=1$.
On the other hand, $\displaystyle f$ attains its inferior value as $\displaystyle x\to\infty$ (as you have shown).
For the second one, let
$\displaystyle g(x):=x+\frac{1}{x}$
for $\displaystyle x>0$ (I believe it must be given on the positive halfline).
Clearly, as $\displaystyle x\to0$ or $\displaystyle x\to\infty$, we have $\displaystyle g(x)\to\infty$ (superior value).
And on the other hand, we have
$\displaystyle g^{\prime}(x)=1-\frac{1}{x^{2}},$
which implies $\displaystyle g^{\prime}(1)=0$
and thus $\displaystyle g(x)\geq g(1)=2$ for all $\displaystyle x>0$ (minimum value).
You may similarly show the third one too, I have to say that this one is very interesting, just show that its maximum values decrease and similarly show that its minimum values increase.
$\displaystyle \lim_{x\to\infty}\ln(x)=\infty$ implies $\displaystyle \lim_{t\to\infty}\frac{1}{1+(\ln(x))^{2}}=0$ which is the inferior value, because there is no $\displaystyle x_{0}$ point that $\displaystyle \frac{1}{1+(\ln(x_{0}))^{2}}=0$ holds.
As I said before $\displaystyle x+\frac{1}{x}$ is unbounded; i.e., its superior value is $\displaystyle \infty$.
And it takes its minimum value at $\displaystyle x=1$ with $\displaystyle 1+\frac{1}{1}=2$.
Note. To find the extreme points of a function $\displaystyle f$, just find the zeros of the equation $\displaystyle f^{\prime}=0$.
Say $\displaystyle x_{0}$ is such a point that $\displaystyle f^{\prime}(x_{0})=0$, then if $\displaystyle f^{\prime\prime}>0$ we say that $\displaystyle f$ has a minimum point at $\displaystyle x_{0}$, and if $\displaystyle f^{\prime\prime}<0$ we say that $\displaystyle f$ has a maximum value at $\displaystyle x_{0}$.
PS. I just now saw that it seems hard by proving in the way I told you before.
An obvious solution is x = 0. The function $\displaystyle f(x) = \frac{\sin x}{x}$ has a removable singularity at x = 0. It makes sense to define f(0) = 1.
Then the global maximum of $\displaystyle f(x) = \frac{\sin x}{x}$ is located at (0, 1).
The other local maxima (and local minima) cannot be found algebraically. You will need to use a numerical procedure to get approximate solutions.
Yes it is not easy, thus, we have to find other ways to find max and min values of this function.
May the the following analysis may be helpful to you.
Let $\displaystyle f(t):=\sin(t)/t$ for $\displaystyle t\in[0,\infty)$.
Clearly, for all $\displaystyle t\in[1,\infty)$, we have $\displaystyle f(t)\leq|\sin(t)|\leq1$.
Thus, if there exists a point $\displaystyle t_{0}\in[0,1]$ such that $\displaystyle f(t_{0})\geq1$ holds, then we can say that $\displaystyle f(t_{0})$ is the maximum value of $\displaystyle f$ on $\displaystyle [0,\infty)$.
By calculating $\displaystyle f^{\prime}(t)=\big(t\cos(t)-\sin(t)\big)/t^{2}$, which implies $\displaystyle f^{\prime}(0)=0$, therefore $\displaystyle f(0)=1\geq1$ is its maximum value.
I saw mr fantastic's arguments after posting the answer.