Limit Help Please

[ChexPex]

Lim as x approaches 0^+ (has that positive sign at the top) of: square root of x times e^(sin (pi/x)) is equal to 0 (We have to prove it)

Anyone got any ideas?

[Peaches09]

x^(1/2)= something, i forgot the derivative formula for that, e^whatever =0, somethingx0=0

[Click]

print this and paste it as your proof.

P.S. This should be in general section.

[rigor]

Please move and/or delete this thread. I have NO desire to come to the forum and see posts about math help.

[austinthecity]

yea.. use the squeeze theorem..

try the parameters:

1/e <= e^sin(pi/x) <= e

[anathema]

looks like you just have to find out what e^sine(180/0) is, so someone prove that equals any number. any number multiplied by x = 0.

this was soo long ago.

[austinthecity]

not quite... you can't divide by 0.

[anathema]

wat, that's a 0+ meaning from the right side

[anathema]

sooo, in terms of english, sine will oscilate between -1 and 1, as it approaches infinity meaning e^-1 all the way through e^1, meaning all those numbers multiplied by x, as x approaches to 0 equals 0. then square root of 0 = 0. voila!

now as for the written proof, no idea Edited October 6, 2008 by anathema

[austinthecity]

in order to prove it, you have to use the squeeze theorem... you can't just say sqrt(x) goes to 0 and it's multiplied to the whole limit goes to 0. the limit may not exist...

for squeeze thm:
[i'm going to use "<" but it's really less than or equal to]
we know

-1 < sin(pi/x) < 1

so
e^-1 < e^sin(pi/x) < e^1

so
sqrt(x)*e^-1 < sqrt(x)*e^sin(pi/x) < sqrt(x)*e

and since we know the limit of the outside functions -> 0, it "squeezes" our function to 0 as well.

[anathema]

blah blah in short 0 times sin(x) as x goes to infinity equals 0. never really liked to show work

pretty pro solve nonetheless Edited October 6, 2008 by anathema