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yaron (eyaron) wrote:

:   lim(1+ax+bx^2)^(-2x)=sqr(e)   (x goes to infinite)

You must have mistyped this problem.
So long as either 'a' or 'b' is non-zero, (1+ax+bx^2) becomes large (in 
absolute value) as x goes to infinity.  So a (large) negative power of 
this number will approach ZERO.

 :   lim (1-x)^ln(x)               (x goes to 1)

I hope you meant to say 'x goes to 1 from below'.
This expression has problems for x bigger than one!

Method 1:  Get out your calculator and calculate the value for x=.9 then 
x=.99 then x=.999 then ...

Method 2:  Look in your calculus textbook under "L'Hopital's Rule"
There should be a section describing what to do with 0^0.

The standard trick is to turn it into and exponential with a fractional 
exponent and use standard L'Hopital on that exponent.
That's enough hinting.
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