Dear ajit pal

y= Lt _x→o  x^x

log y = Lt _x→o  xlog x

        =Lt _x→o  (log x)/(1/x)

        =Lt _x→o  (1/ x)/(-1/x²)

        =Lt _x→o  (- x)

        =0

y=e⁰

y=1

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lim x^x
x -> 0
I'm not sure this limit exists unless the limit is one-sided. I'll assume it to be one-sided.
lim x^x
x -> 0+

Use the property y = e^(ln(y)). Since e and natural log are inverses of each other, they cancel each other out.

lim e^(ln(x^x))
x -> 0+
Use the log property to move the x outside of the log.

lim e^[x ln(x)]
x -> 0+
Move the limit inside of the exponent.

e^[ lim (x ln(x)) ]
. . x -> 0+
Now the limit is of the form [0*infinity], which is indeterminate. Let's move x to the denominator as (1/x).

e^[ lim ( ln(x)/(1/x) ) ]

. . x -> 0+
How the form is [infinity/infinity]. Use L'Hospital's rule.
The derivative of ln(x) is (1/x). The derivative of (1/x) is (-1/x^2).

e^[ lim ( (1/x)/(-1/x^2) ) ]
. . x -> 0+

Simplify.

e^[ lim ( (-x^2)/x ]
. . x -> 0+

e^[ lim (-x ) ]. . x -> 0+

And evaluate the limit
e^0 Which is just 1

--

regards

Ramesh