Karl's Calculus Tutor - Solution to Exercise 8.1-5

Solution to Exercise 8.1-5KCT logo

© 1998 by Karl Hahn

The problem was to find the limit of 

    lim     xx
   x  > 0+
and the hint was that xx = ex ln(x), and that if you can find the limit of x ln(x), then you can easily find the exponential of that limit. Note that because ex is a continuous function, 
    lim  ex ln(x)  =  elim x ln(x)
which is why we need only find the limit of x ln(x).

Step 1: Write it as a quotient. The thing we are taking the limit of is, right now, expressed as a product. We need to write it as a quotient. 

                                 ln(x)
    lim     x ln(x)  =   lim          
   x  > 0+              x  > 0+     1
                                    
                                   x

Step 2: What do numerator and denominator go to in the limit? The denominator, 1/x, clearly grows without limit as x approaches zero -- that is 1/x goes to infinity. The function, ln(x) is not defined at x = 0, but it does grow without limit in the negative direction as x approaches zero -- that is it goes to minus infinity. So our new version of L'Hopital's Rule applies.

Step 3: Find the derivatives the numerator and denominator. We know that the derivative of ln(x) is 1/x. And the derivative of 1/x is -1/x2.

Step 4: Apply L'Hopital's Rule. Which means that 

                                 1
                                  
            ln(x)                x
    lim            =   lim          
   x  > 0+    1       x  > 0+    -1
                                   
              x                  x2

Step 5: Take the limit. First multiply top and bottom by -x2. Now you have 

            ln(x)
    lim            =   lim     -x
   x  > 0+     1       x  > 0+
               
              x
The limit as x approaches zero of -x is zero. So you end up with 
            ln(x)
    lim            =  0
   x  > 0+     1
               
              x
Now recall that what we were really looking for was the limit as x approaches zero of ex ln(x). And from the above calculations, that limit would have to be e0 = 1. It must also be true, then, that 
    lim     xx  =  1
   x  > 0+
If your calculator has an exponentiation function, you can try raising a small positive x like 0.0000001 to itself and see for yourself what the trend is.

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