What is the limit as x approaches 0+of e^{x^2ln(x)} ?

The limit as x approaches 0+of e^{x^2ln(x)} is 1

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limit as x approaches 0+of e^{x^2ln(x)}

x \to 0 + lim (e^{x^{2} l n (x)})

1

Solution steps

x \to 0 + lim (e^{x^{2} l n (x)})

Simplify

e^{x^{2} l n (x)} : x^{x^{2}}

= x \to 0 + lim (x^{x^{2}})

Apply exponent rules

= x \to 0 + lim (e^{x^{2} l n (x)})

Apply the Limit Chain Rule:

1

= 1

h \to 0 lim (\frac{2 x + 2 h + 1 - 2 x + 1}{h})

x \to 3 lim (\frac{1}{( 3 + x ) ^{2}})

x \to 7 lim (x + 7)

x \to 1 + lim (\frac{4}{x ^{2} - 1})

What is the limit as x approaches 0+of e^{x^2ln(x)} ?
The limit as x approaches 0+of e^{x^2ln(x)} is 1