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

The limit as x approaches 0+of e^x+ln(x) is -infinity

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

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

- \infty

Solution steps

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

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

With the exception of indeterminate form

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

x \to 0 + lim (e^{x}) = 1

x \to 0 + lim (ln (x)) = - \infty

= 1 - \infty

Apply Infinity Property:

c - \infty = - \infty

= - \infty

\int \frac{18}{( x + 3 ) ( x ^{2} - 9 )} d x

y^{^{''}} + 2 y^{^{'}} + y = \frac{- 4 e ^{- t}}{t ^{2} + 1}

\frac{\partial}{\partial x} (\frac{y ^{2} - x ^{2}}{x y})

n = 1 \sum \infty \frac{4}{7 n + 2}

\frac{d}{d x} (x^{5} e^{- 3 x})

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