can anyone help me with this limit?
limit (x->0) xx
Hint:
Let \(y={\lim }\limits_{x\to 0} x^x\)
\(\Rightarrow\ln y={\lim }\limits_{x\to 0} x\ln x\)
\(\Rightarrow\ln y={\lim }\limits_{x\to 0} {{\ln x}\over{{1/x}}}\)
Which is \({\infty \over \infty }\) form, hence L. Hospital's rule can be used.
write thx in the base as (1+(x-1))
then multiply and divide the x in the exponent by x-1
so we have
lim x->0 (1+(x-1))x(x-1)/(x-1)
this reduces to the form lim x->0 (1+p)1/p
so we get elim x->0 x(x-1) =1
I think the limit does not exist.
LHL = lim h->0(0-h)^(0-h) =lim h->0(-1)^(-h)xh^(-h)
lim h->0(1/h)^h
RHL=lim h->0(0+h)^(0+h)= lim h->0h^h
Hence LHL is not equal to RHL.
Hence the limit does not exist.