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

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

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limit as x approaches 1+of e^{ln(ln(x))}

x \to 1 + lim (e^{l n (l n (x))})

0

Solution steps

x \to 1 + lim (e^{l n (l n (x))})

Apply log rule:

a^{l o g_{a} (b)} = b

= x \to 1 + lim (ln (x))

Plug in the value

x = 1

= ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

e^{- y}

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n = 1 \sum \infty n^{2} e^{- n^{2}}

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