Gdtyjr

I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$

For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.

For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?

Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$

Without L'Hospital
$$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$

Now, by Taylor for large values of $x$
$$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
$$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
$$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
$$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached

I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.

You can solve the first one using

• $arctan x + operatornamearccotx = fracpi2$


• $lim_yto 0(1-y)^1/y = e^-1$


• $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$

begineqnarray* left(frac2pi arctan x right)^x
& stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
& stackrelx to +inftylongrightarrow & e^-frac2pi
endeqnarray*

The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider

• 
  $fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.