Expert: Paul Klarreich - 10/28/2006

Question
I got a question for a couple of problems I am working.  I want to make sure I am doing this right.
lim [cos(2/x)]^x^2
x->(infinity)

This is how I proceeded: x^2 ln cos 2/x = ln cos (2/x)/(1/x^2) not sure how to proceed from here.

One other one that perplexed me was this one:

lim   [1/x - 1/(e^x - 1)]
x->0

Oh if you didn't figure it I am trying to find the limit.  Thanks for any guidance you can give me.

Answer
Questioner:  Gary
Category:  Calculus
 
Subject:  Exponential, Log and Inverse Trig Functions
Question:  I have a question for a couple of problems I am working.  I want to make sure I am doing this right.
lim [cos(2/x)]^x^2
x->(infinity)

This is how I proceeded: x^2 ln cos 2/x = ln cos (2/x)/(1/x^2) not sure how to proceed from here.

One other one that perplexed me was this one:

lim   [1/x - 1/(e^x - 1)]
x->0

Oh if you didn't figure it I am trying to find the limit.  Thanks for any guidance you can give me.
 
.......................................................
Hi, Gary,

I am going to assume you have learned (or are learning) l'Hospital's rule.  If you have not, let me know and I'll try to come up with something else.

For:

lim  [cos(2/x)]^x^2
x->inf

Let's see -- this looks like something that approaches 1 raised to a high power.  That would normally just be 1, but with these limits, anything can happen.

I am going to assume that the expression should be parenthesized this way:

lim  [cos(2/x)]^(x^2)
x->inf


Try taking the log of the expression:

ln [cos(2/x)]^(x^2) =  x^2 ln cos(2/x)


I think it will be easier and we get the same result using this substitution:

y = 1/x
x^2 = 1/y^2
and  x->inf  means  y->0

    ln cos(2y)
lim  ----------    << that's a bit simpler.
y->0    y^2

I think we can use l'Hospital's rule on this, which means differentiating top and bottom independently:

    1/cos(2y) (-2) sin (2y)
lim  -----------------------
y->0        2y

    -2 sin 2y
lim  ----------
y->0  2y cos 2y

      -2    sin 2y
lim  ------- ------
y->0 cos 2y    2y

Now, as y->0, the first factor becomes -2/1 = -2.
The second factor is a well-known limit; it's equal to 1.

So this limit evaluates to -2.

Therefore the original limit evaluates to e^(-2) = 1/e^2 ~~ 0.135
...........................................................

One other one that perplexed me was this one:

lim   [1/x - 1/(e^x - 1)]
x->0

Any time you have a difference of two things that both approach infinity, try to make it into a single term.  In this case, combine the fractions:

1        1
--- -  -------    LCD = x(e^x - 1)
x     e^x - 1

e^x - 1 - x
------------
x(e^x - 1)

Now this is in the form  0/0, and we can use l'Hospital's rule again.

   e^x - 1
---------------------
(1)(e^x - 1) + x(e^x)

   e^x - 1
---------------
e^x - 1 + xe^x

Still 0/0.  But we're getting there.  (Hmmm.. my dentist said the same thing, half way through pulling that tooth.)

One more application of l'Hospital's rule:
     e^x
-----------------
e^x + e^x + xe^x
    1
----------
1 + 1 + x

Ok, that limit as  x->0 is easy.  It's 1/2.