Expert: Paul Klarreich - 6/20/2008

Question
I hope you can help! I am taking a college first year calculus correspondence class. I have am confused with the following three problems and am really hoping to find out how to work limit problems like these! If you could explain how to obtain the correct answer it would be greatly greatly appreciated!!!  

Directions: Find the limit. Use L'Hospital's Rule where appropraite. If there is a more elementary method, consider using it. If L'Hospital's Rule doesn't apply, explain why.

Q1: The limit as x approaches infinity of lnlnx divided by x
Q2: The limit as x approaches negative infinity of x^2 e^x
Q3: The limit as x approaches 0 from the right of (tan2x)^x

On Q1 I'm pretty sure that lnlnx/x becomes (1/xlnx)/1 but I'm stuck after that point. I hope you can help with these!!! Thanks.

Answer
Questioner:   Lee
Category:  Calculus
Private:  No
 
Subject:  Calculus I (Indeterminate Forms and L'Hospital's Rule)
Question:  I hope you can help! I am taking a college first year calculus correspondence class. I have am confused with the following three problems and am really hoping to find out how to work limit problems like these! If you could explain how to obtain the correct answer it would be greatly greatly appreciated!!!  

Directions: Find the limit. Use L'Hospital's Rule where appropraite. If there is a more elementary method, consider using it. If L'Hospital's Rule doesn't apply, explain why.

Q1: The limit as x approaches infinity of lnlnx divided by x
Q2: The limit as x approaches negative infinity of x^2 e^x
Q3: The limit as x approaches 0 from the right of (tan2x)^x

On Q1 I'm pretty sure that lnlnx/x becomes (1/xlnx)/1 but I'm stuck after that point. I hope you can help with these!!! Thanks.
......................................
Hi, Lee,

Q1: The limit as x approaches infinity of lnlnx divided by x

When you want to use L'Hospital's rule, you need a limit that:

1. Is in the form of a quotient.
2. Produces 0/0 or inf/inf.
      ln(ln x)   << use () here to make it clear.
lim    --------
x->inf    x

     ln(ln inf)     inf
--->  ---------- --> ---
         inf        inf

So far, so good.  Now apply the rule and differentiate.

(1/ln x)(1/x)     1
------------- = ------
    1          x ln x
            1
That gives ----------- = 0
          inf ln(inf)

End of story.
............................
Q2: The limit as x approaches negative infinity of x^2 e^x
             
Lim     x^2 e^x
x->-inf

--> (-inf)^2 e^(-inf) = inf * zero.

Sort of OK, but you want a quotient to apply the rule.  Write:
              x^2
lim         --------
x-> - inf     e^-x

Doesn't look nice, but YOU HAVE TO HAVE A QUOTIENT, so you apply Clint Eastwood's rule. [Man's gotta do what a man's gotta do.]

Now differentiate:

 2x
-------
- e^-x
Apply the limit:
          2x
lim     ------ -->
x->-inf  - e^-x

-2(inf)
--------
e^(+inf)

Still N.G. but you apply the rule a second time:

 2
-------
+e^-x

Apply the limit:
          2         2
lim     ------ --> ------ = 0
x->-inf  e^-x      e^inf

Next.
...............................

Q3: The limit as x approaches 0 from the right of (tan2x)^x

I assume you mean:
(tan(2x))^x   << don't forget the ().

Try this:

Let  y = ln((tan(2x))^x) -- a good thing to do any time you have some exponential function.  When you get your  lim y, you just do e^(answer).


y = ln((tan(2x))^x)

lim y = lim ln((tan(2x))^x)

= lim  x ln(tan(2x)))

Now as  x --> 0+, that is  0 ln (tan(0))

= 0 ln (0) = 0 * - inf, which needs work as usual.

{yes, remember that ln 0 is - inf.}

Write  
ln(tan(2x))
------------
  1/x

Differentiate:

1/tan(2x) * sec^2(2x)(2)
---------------
   -1/x^2


2 x^2 * sec^2(2x)
-----------------
   -tan(2x)

Now sec(0) = 1, so we can ignore that factor.  [I.E. that is:


-2 x^2   
--------- sec^2(2x)
tan(2x)

The lim of the second factor is 1, so forget it.
-2 x^2    -  2x     x cos 2x
------- = -------- -------------
tan(2x)   sin(2x)       1

= - 1 * 0 = 0

So y = 0, and the answer is  e^0 = 1.

Fun.