I need some help with these problems.

Evaluate

1) ${\lim }_{x\to \pi /4}(\tan 2x{)}^{\tan x}$

( $x$ in radians)

2) evaluate integral of

$[x^2]\mathrm{dx}/[(x-14{)}^2]+[x^2]$ from 0 to 14

Hi Archishman,

Sorry for delay, I've been away.

(1) I will look at the limit $x\to \pi /4$ with $x<\pi /4$. This is because for $x$ slightly greater than $\pi /4$, $\tan (2x)<0$ and then we have a negative number to the power of $\tan (x)$. We could look at that later, but discuss just limit from below for now.

$\tan (2x)\to \infty$and $\tan (x)\to 1$ as $x\to \pi /4$. This indicates that $(\tan (2x{))}^{\tan (x)}$ also tends to $\infty$.

To show this rigorously, restrict to $x$ suitably near $\pi /4$ that $\tan (x)$ is greater than 1/2 then note $\tan (2x{)}^{\tan x}>\tan (2x{)}^{1/2}$. The right hand side shoots off to $\infty$ as $x$ approaches $\pi /4$, hence the left hand side does likewise.

(2) This is your integral? ${\int }_0^{14}[x^2]/([(x-14{)}^2]+[x^2])\mathrm{dx}$

Set,

$I={\int }_0^{14}[x^2]/([(x-14{)}^2]+[x^2])\mathrm{dx}$

$J={\int }_0^{14}[(x-14{)}^2]/([(x-14{)}^2]+[x^2])\mathrm{dx}$

I think $I=J$ by setting $y=x-7$ in $I$ and $y=7-x$ in $J$. Note also $I+J={\int }_0^{14}\mathrm{dx}=14$.

So $I=7$.

Dear Ian,

I am so sorry, I actually typed the wrong limit problem. The problem that I wanted to give is

${\lim }_{x\to \pi /2}\tan x^{\tan 2x}$

Well that's okay. Here are the essentials:

# Once again I'll look at the real limit from below, that is $x<\pi /2$ for the same reason as before.

# Set $y=\backslash p/2-x$ so that we are looking for the limit of $\tan (y{)}^{\tan (2y)}$ as $y$ approaches 0 from above.

# Roughly speaking, $\tan (y)~y$ and $\tan (2y)~2y$ near the origin so the limit is essentially that of $y^{2y}$ which is 1.

# $y^{2y}=e^{2y\log (y)}$ and $y\log (y)\to 0$ as $y\to 0$ from above (need any help showing this?)

# To elaborate on the 'roughly speaking' part you could show that for $a$, $b>0$, $(\mathrm{ay}{)}^{\mathrm{by}}\to 1$ as $y$ tends to 0 from above and sandwich $\tan y^{\tan 2y}$ between two such limits. E.g. $y^{4y}<\tan y^{\tan 2y}<(2y{)}^{2y}$ for small enough positive $y$.

Is that enough? Is that clear? Tell me if any problems,

Ian,

forgive me if I am being stupid ,but when you put $y=\pi /2-x$, $y\to 0$ so $y^{2y}$ is something like $0^0$.

Hi!

# Archishman: The line after I first wrote $y^{2y}$ is meant to explain why the limit is 1. I'll be more explicit here:

For POSITIVE $y$, $y^{2y}=e^{2y\log y}$. Now $y\log y\to 0$ as $y\to 0$ ((set $y=e^{-z}$ then $y\to 0$ equivalent to $z\to \infty$and $y\log y=-z/e^z\to 0$ )) Hence $y^{2y}\to e^{2\times 0}=1$.

Tell me how you go......