I need some help with these problems.

Evaluate

1) lim_x®p/4 (tan 2x)^{tan x}

(x in radians)

2) evaluate integral of

[x²]dx/[(x-14)²] + [x²] from 0 to 14

Hi Archishman,

Sorry for delay, I've been away.

(1) I will look at the limit x ®p/4 with x < p/4. This is because for x slightly greater than p/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) ®¥and tan(x) ®1 as x ®p/4. This indicates that (tan(2x))^tan(x) also tends to ¥.

To show this rigorously, restrict to x suitably near p/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 ¥ as x approaches p/4, hence the left hand side does likewise.

(2) This is your integral? ò₀¹⁴[x²]/([(x-14)²]+[x²])dx

Set,

I = ò₀¹⁴[x²]/([(x-14)²]+[x²])dx

J = ò₀¹⁴[(x-14)²]/([(x-14)²]+[x²])dx

I think I=J by setting y=x-7 in I and y=7-x in J. Note also I+J=ò₀¹⁴dx=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®p/2tan 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 < p/2 for the same reason as before.

# Set y=\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) ®0 as y ®0 from above (need any help showing this?)

# To elaborate on the 'roughly speaking' part you could show that for a, b > 0, (ay)^by ®1 as y tends to 0 from above and sandwich tan y^{tan 2y} between two such limits. E.g. y^4y < tan y^tan2y < (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=p/2-x, y® 0 so y^2y is something like 0⁰.

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 ® 0 as y ® 0 ((set y=e^-z then y ® 0 equivalent to z ®¥and y log y = -z/e^z ® 0 )) Hence y^2y ® e^2×0 = 1.

Tell me how you go......