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

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

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

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

0

Solution steps

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

Simplify

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

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

Plug in the value

x = 0

= e^{2} \cdot 0

Simplify

= 0

x \to 7 - lim (f (x))

n = - \infty \sum \infty 1

n = 0 \sum \infty e^{- n}

arctan (\frac{3 + t}{3 - t})

x \to 0 + lim (x^{7 s i n (x)})

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