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

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

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

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

0

Solution steps

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

Apply log rule:

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

= x \to 0 + lim (x)

Plug in the value

x = 0

= 0

x \to a lim (f (x) - g (x))

x \to 2 - lim (9^{\frac{( - 1 )}{x}})

x \to 0 lim (csc (x) + \frac{1}{x})

x \to \infty lim (5 x e^{x})

x \to 3 - lim (\frac{3 x}{x - 3})

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