$$\lim_{x \to 0} x^{x}$$​ is equal to:

Correct option is B

Given:
​$$\lim_{x \to 0^{+}} x^{x}$$​​
Formula used:
​$$x^{x} = e^{x \ln x}$$​​
and
​$$\lim_{x \to 0^{+}} x \ln x = 0$$​​
Solution:
Let $$y = x^{x}$$​​
Taking logarithm:
​$$\ln y = x \ln x$$​​
Now take limit as $$x \to 0^{+}:$$​​
​$$\lim_{x \to 0^{+}} x \ln x = 0$$​​
So,
​$$\ln y = 0\\y = e^{0} = 1$$​​
Hence,
​$$\lim_{x \to 0} x^{x} = 1$$​​
The correct answer is (b) 1.