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

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

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limit as x approaches 0 of ln(x^2)(1e^x+1)

x \to 0 lim (ln (x^{2}) (1 e^{x} + 1))

diverges

Solution steps

x \to 0 lim (ln (x^{2}) (1 \cdot e^{x} + 1))

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

x \to 0 + lim (ln (x^{2}) (1 \cdot e^{x} + 1)) = - \infty

x \to 0 - lim (ln (x^{2}) (1 \cdot e^{x} + 1)) =

Does not exist

2

= diverges

\frac{d}{d t} (3 cos^{3} (t))

3^{(- \frac{1}{10} t)}

\frac{- 5 e ^{(- 3 s)}}{( s ^{2} - 5 s + 6 )}

\int \frac{2 x ^{2} - 5}{x ^{4} - 5 x ^{2} + 6} d x

\int (2 x^{2} + 2 x - 3)^{10} (2 x + 1) d x

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